
LIBRARY OF CONGRESS. 

Chap, A__„. Copyright No.. __. 



Shelf..,M.^.? 



<;/>-. 



UNITED STATES OF AMERICA. 



Copyright 1897. 
THOMAS W. MATHER. 



A TEXT-BOOK 



OF the; 



Strength of M.^'^'erials 



AND OF 



STRESSES IN STRUCTURES 



FOR USK IN 

SCIENTIFIC HIGH SCHOOLS AND SCHOOLS OF 

TECHNOLOGY. 



BY 



THOMAS W. MATHER, Ph. B., M. E. 

PRINCIPAL OF THF BOARDMAN SCIENTIFIC HIGH SCHOOL, 
NEW HAVEN, CONN. 




NEW HAVEN, CONN.: 
BOARDMAN SCHOOL PREvSS, 

1897. 

u 







l^ 



PREFACE. 

A study of the Strength of Materials and of methods of 
finding the stresses which occur in structures, has hitherto 
been usually left to Schools of Technology and Colleges. 
The reasons for this appear to be, either that all of the 
time in the lower school is needed to prepare for entrance 
examinations into the higher; or it is conceived that the 
mind can be better trained in some other way; or that this 
and allied subjects are too difficult to be successfully taught 
in High and Manual Training Schools. 

In case the pupil is expecting to enter a college, it is of 
course necessary to meet the preparatory requirements, for 
which most of the available time may be needed, but the 
Author is strongly of the opinion that in schools from which 
the great majority pass directly into the active work of life, 
such studies should occupy a prominent place; that, while 
the purel}^ mathematical studies, Algebra, Geometry and 
Trigonometry are essential, they should be looked upon as 
tools which the pupil must be taught to use; and that 
through practical application, mental development will be 
much more more marked and lasting. 

Experience has convinced the Author that this applica- 
tion is entirely within the capacit}^ of High School boys, 
and for the purpose of providing a suitable text-book, he 
has compiled this little volumn, to be followed, he hopes, 
by others of a similar nature. In order to make room, he 
is of the opinion that the course in pure mathematics may 



IV 

profitably be shortened, especially in Algebra beyond 
Quadratics, and in Solid and Spherical Geometry; but in 
the study of Applied Mathematics, an elementary knowl- 
edge of the Calculus is almost a necessity, many deductions 
otherwise being unintelligible. Only the elements are 
required, and these the Author believes can be much more 
readily mastered than many of the propositions in Higher 
Algebra and Solid Geometry, and will certainly prove 
vastly more comprehensive and useful. Accordingly, in 
the school over which he presides, a short preparatory 
course is given. 

In the following pages however, excepting Chapters III 
and VII, the use of Calculus has been avoided, without 
sacrifice it is hoped, of clearness. Where emplo3^ed, the 
student not familiar with the method, must accept the re- 
sults on faith. 

The subject is one which readily lends itself to experi- 
mental demonstration, and interest in the w^ork and its 
value to the student wdll be in this way greatly increased. 

The Author is indebted to many writers and investiga- 
tors, in compiling this volume, especially to Prof. Mansfield 
Merriman, to whose admirable treatises on " The Mechanics 
of Materials" and "Roofs and Bridges" the student is 
referred for a more complete treatment of the subjects. 

THOS. W. MATHER. 
Nkw Haven, Jan., 1897. 



CONTENTS. 



CHAPTER I. 
MATERIALS OF CONSTRUCTION. 

PuRK Iron. Iron and Carbon. Cast Iron. Othkr 
A1.L0YS OF Iron. Copper and its Ai,i,oys. Average 
Weights. 

CHAPTER II. 

TENSILE, COMPRESSIVE, AND SHEARING 

STRESSES. 

Definitions. Experimental Laws. Tabee of 
Uetimate Strength. Tabee of Unit Stresses at 
THE Eeastic Limit. Coefficients of Elasticity. 
Elongation and Compression. Elongation of a 
Rod under its own Weight. The Resilience of 
Materials. Tensile and Compressive Stresses. 
Factor of Safety. Table of Factors. Shearing 
Stresses. Stress Produced by Change of Tem- 
perature. 

CHAPTER III. 

STRESSES IN AND STRENGTH OF BELTING. 

Fundamental Equations. Transmission of Power. 
Strength of Leather Belting. Rope Belting. 
The Strength of Ropes. Influence of Centrifu- 
gal Force. 



VI 



CHAPTER IV. 

CYLINDERS, RIVETED JOINTS, BOILERS. 

Fundamental Equations. Pressure Caused by 
Liquids. Thick Cylinders. Flues. Riveted Joints. 
Boiler vShells. Table of Dimensions and Effi- 
ciency. Strength of Boilers. 

CHAPTER V. 

BEAMS AND CANTILEVERS. 

Reactions on the Supports. Vertical Shear. 
Bending Moments. Maximum Bending Moments. 
General Expressions for the Bending Moment 
AND Vertical Shear. 

CHAPTER VI. 
THE STRENGTH OF BEAMS. 

The Neutral Surface and Neutral Axis. Funda- 
mental Equations. Centre of Gravity. The Re- 
sisting Moment. Moments of Inertia. Tables. 
Graphical Method. Modulus of Rupture. The 
Stresses in Beams. Beams of Uniform Strength. 

CHAPTER VII. 

THE DEFLECTION OF BEAMS. 

Equation of the Elastic Curve. Cantilevers. 
Simple Beams. Restrained Beams. 



Vll 



CHAPTER VIII. 
STRESSES IN AND STRENGTH OF SHAFTS. 

The Twisting Moment. The Resisting Moment. 
P01.AR Moment of Inertia. Transmission of Power. 

CHAPTER IX. 
COMBINED STPvESSES. 

Long Coeumns. Radius of Gyration. Rankine's 
Formula. Practical Applications. Compound 
Columns. Combined Tension and Flexure. Com- 
bined Tension or Compression and Shear. 
Combined Torsion and Flexure. Combined Tor- 
sion AND Compression. 

CHAPTER X. 
STRESSES IN ROOF AND BRIDGE TRUSSES. 

Definitions. Reactions. Moments. Vertical 
Shear. Determination of Stresses. Stresses in a 
Bridge Truss. Dead Loads on Roof Trusses. Dead 
Loads on Highway Bridges. Dead Loads on Rail- 
road Bridges. Practical Applications. Stresses 
DUE TO Wind Pressure. Pressure Caused by Wind. 
Stresses due to a Single Moving Load. Stresses 
DUE TO Live Loads. Live Loads on Highway 
Bridges. Live Loads on Railroad Bridges. The 
Length and Depth of Panels. Application to a 
Pratt Truss. Double or Multiple Systems. The 
Bowstring Girder. Kinds of Roof and Bridge 
Trusses. 



THE STRENGTH OF MATERIALS 

AND 

STRESSES IN STRUCTURES. 



CHAPTER I. 



MATERIALS OF CONSTRUCTION. 
Art. I. PuRK Iron. 

Chemically pure iron is never met with in practice; even 
the purest wrought iron containing a noticeable amount of 
carbon with traces of other minerals. It may however be 
readily produced in the laboratory, and, as with many of 
the elements, its physical properties vary with its manner 
of production. In one case it is in the form of a dark gray 
powder which combines energetically with oxygen, taking 
fire spontaneously when thrown into the air. In another 
there results a brilliant button of metal exhibiting a de- 
cidedly crystaline structure. This is the usual appearance 
of pure iron. 

According to Dana (Minerology) it crystalizes in the 
isometric system, and has a specific gravity varying from 
7.3 to 7.8. It is susceptable of being magnetized to a much 
higher degree than steel, but unlike the latter, does not 
retain its magnetism when the exciting cause is removed. 
Its specific heat is .1138 (Regnault), and according to 
Despret its conducting power is .374, gold being taken 



as i.oo. Its electric resistance as determined by Poullet is 
5.88 times that of a copper conductor of equal sectional 
area. 

Art. 2. Iron and Carbon. 

* Of all the compounds of iron none can be compared 
with those of carbon in practical importance and, from a 
scientific point of view, none possess greater interest. The 
influence of the element in causing variations in the ph^^si- 
cal properties of iron is one of the most extraordinar}' 
phenomina in the whole range of metalurgy. 

Under the common name of iron are included virtually 
distinct metals which in external characters differ far more 
from each other than many chemically distinct. Without 
carbon the manifold uses of iron would be greatl}'^ re- 
stricted: and so far as it is known no other metal or 
mixtures of metal can be applied to these uses. 

When carbon is absent or only present in very small 
quantity, v,^e have Wrought Iron, which is comparatively 
soft, malleable, ductile, weldable, easily forgeable and 
very tenacious, but not fusible except at temperatures 
rarely attained in furnaces, and not susceptable of tem- 
pering. 

When present in certain proportions, the limits of which 
cannot be exactly prescribed, we have the various kinds of 
Steel, which are highly elastic, malleable, ductile, forge- 
able, weldable, capable of receiving very different degrees 
of hardening b}" tempering, and fusible in furnaces. 

'L/Sistly when present in greater proportion than in steel 
we have Cast Iron, which is hard, comparatively brittle, 
and readily fusible but not forgeable or weldable. 

The differences between these well known sorts of iron 



* Percy's Metalurgy. 



essentially depend upon differences in the proportion of 
carbon, though other elements may and often do concur in 
modif3dng in a striking degree the qualities of this wonder- 
ful metal. 

The transitions from Wrought Iron to Steel and from 
Steel to Cast Iron are so gradual and so insensible that it 
is impossible to pronounce when one begins and the other 
ends. Karsten has made the following empirical classifi- 
cation: made under the supposition that very small amounts 
of foreign matter, other than carbon are present. 

TABLE I. 

Name. Per Cent, of Carbon. Properties. 

1. Malleable Iron 25 and less Is not sensibly hard- 

ened by cooling. 

2. Steely Iron .35 Can be slightly hard- 

ened by cooling. 

3. Steel 50 Gives sparks with a 

flint when hardened. 

4. " I. GO to 1. 50 Limits for steel of 

max. hardness and 
tenacity. 

5. " 1.75 Superior limit of 

welding. 

6. " 1.80 Very hard Cast Steel, 

forging with great 
difficulty. 

7. " 1.90 Not malleable hot. 

8. Cast Iron 2.00 Lower limits of cast 

iron. Cannot be 
hardened. 

9. " " 6.00 Highest carburetted 

compound attainable. 



Art. 3. Cast Iron. 

. The carbon existing in Wrought Iron and Steel cannot 
be distinguished bj^ the eye, being combined with the metal, 
but in Cast Iron it also occurs in what is known as the 
"graphitic state." In this condition the fracture is dark 
gray, granular or scaly crystaline, and the iron is known as 
Gray Iron. When, on the other hand, the carbon is mostly 
combined, the fracture is more or less white, granular or 
crystaline, and it is known as White Iron. 

The gray iron passes by insensible gradations into white, 
and at certain stages, both kinds are visible in the fractured 
surface, constituting what is called Mottled Iron. From 
the extreme of grayness to absolute whiteness not less than 
eight varieties are recognized, and these are designated by 
the numerals from one to eight respectively. 

The mode of existence of the carbon depends in great 
measure upon conditions of solidification in cooling from a 
melted state; and upon the temperature at which fusion 
was effected. Rapid solidification favors the retention of 
carbon in the combined state, and by this means it is pos- 
sible to convert gray iron into the perfectlj^ white. Thus 
by pouring liquid gray cast iron into a cold metalic mould, 
so as to cause sudden cooling, the exterior where it comes 
in direct contact with the mould will de found to be in the 
state of white iron, while the interior will remain gray. 
This principle is extensively employed in practice in the 
process known as chill casting. 

Gray iron has a higher melting point than white, and 
in fusing passes almost instantly from the solid to the 
liquid state, where it is very fluid; whereas white iron at 
lower temperatures becomes first soft then pasty before 



melting. The total amount of carbon may be the same in 
both gray and white cast iron, but generall}" the former 
contains the larger quantit3^ 

Art. 4. Other Alloys of Iron. 

Iron readily combines with man}^ of the elements in 
varying proportions, but as a rule the effect is injurious to 
strength, and the introduction is avoided when possible, 
excepting for a special purpose such as increasing fluidity 
or exceptional hardness. The presence of a small amount 
of nickel how^ever, is decidedly beneficial in some cases, as 
in steel armor-plates, and aluminum in cast iron appears 
to. give sounder castings without sacrifice of strength. 

The varieties of w^rought iron, steel, and cast iron com- 
merciall}^ emploj^ed are very great, depending not only 
upon the carbon contained, but also upon the process of 
manufacture. For names, peculiarities and especial object 
of the different grades, the student is referred to engineer- 
ing manuals and trade catalogues. 

Art. 5. Copper and its Alloys. 

Next to iron, copper and its alloys are undoubtedly the 
most important of metals used in construction. Combined 
with tin the alloy is knovv^n as bronze, with zinc it is called 
brass, and since the combination may be in any proportion 
and include many other metals, the number of alloj's dis- 
tinct in many waj^s from each other are very numerous. 
The following table gives the composition by Vv'-eight of 
some of the more iniportent. 



TABLE II. 

Name. Copper. Tin. Zinc. 

Bronze Medals 92 8 

" Gun Metal 90 10 

*' Axle Bearings 84 16 

Bell Metal 78 20 

" Mirror " 66 22 

Brass Pinchbeck 80 20 

" Common 75 25 

" Wire 67 33 

" White 10 10 80 

Aluminum Bronze 89.5 Aluminum 10.5 

Fluxed with phosphorus the alloys of tin and copper are 
known as phosphor bronze and by the process the strength 
and general usefulness is materiall}^ increased. By the 
addition of a small per cent, of manganese a similar effect 
is produced, and these alloys are extensively used. 

Copper also combines readily in all proportions with 
aluminum forming a new series of alloys which are remark- 
able for beauty, lightness and strength: combined with tin 
(.89) and antimony (.7) we have a most valuable material 
for bearings known as Babbitt Metal: combined with zinc 
(.34) and nickel (.34) we have German Silver: with tin 
(.77), antimony (.15) and zinc (2), there results Britannia 
Metal. 

Many other combinations have names and are used for 
especial purposes. The student is referred to Thurston's 
" Materials of Construction " and to Kent's and other 
hand-books for detailed information. 



Art. 6. AvKRAGK Wkights. 

The following table gives the average weights of the 
prii^cipal materials used in construction. 



TABI.K III. 

specific Weight, Weight, 

Gravity. Cub. Inch. Cub. Ft. 

Cast Iron , 7.21 261 

Steel 7.81 282 

Wrought Iron 7.79 278 

Copper 8.88 321 

Gun Metal 8.70 315 

Brass (Common) 8.53 308 

Aluminum 2.56 093 

White Oak 0.86 53.7 

" Pine 0.55 34.6 

Yellow Pine 0.66 41.2 

Beech 0.85 53.2 

Granite 2.64 165.0 

Sandstone 2.20 137-5 

Brick 1.90 1 18.7 

It is useful to remember that a bar of wrought iron one 
yard long and one square inch in section weighs nearly 10 
pounds. 

Prob. I. What should be the cross section in square 

inches of a bar of aluminum in order that it may equal in 

weight a bar of wrought iron of the same length and one 

inch in diameter? Ans. 2.35 sqr. inches. 

3 



8 

Prob. 2. Find the weight of a wrought iron pipe 6 ft. 
long whose diameters are 2 and 2>2 inches. 

Ans. 35.4 lbs. 

Prob. 3. A cast iron pipe 3 ft. long weighs 36.84 lbs. 
What must be its cross section in sqr. inches ? If the oiit- 
side diameter is 3 inches, what must be the inside diame- 
ter ? Ans. 3.93 sqr. inches. 2 inches. 

Prob. 4. A heavy 10 inch wrought iron I beam weighs 
45 lbs. per foot. What must be its cross section in sqr. 
inches ? Ans. 13.5 sqr. inches. 



CHAPTER II. 

TENSILE, COMPRESSIVE, AND SHEARING 

STRESSES. 

Art. 7. Definitions. 

The molecules of a solid body are held in a definite 
position with reference to each other by molecular forces 
which must be overcome whenever the shape of the body is 
changed. The internal resistance which material opposes 
to any external force tending to change its shape is called 
a stress. The tendency of an external force to separate the 
molecules is resisted by their cohesion or mutual attrac- 
tion; and to force them nearer together by the vibrations 
due to heat and other causes which tend to maintain a 
definite molecular distance. 

K unit stress is the stress on a unit of area, and is equal 
to the total stress divided by the area of the material 
opposed to it. Thus let a rod having a cross section of 4 
sqr. in. be pulled by a force of 1000 lbs. The unit stress 
is then I-^-O z=r 250 lbs.; or, generally, if A = the cross 
section in square inches, P = total stress in pounds and 
S = unit stress, we have the relation, 

P 

^ = A '-' 

If the force acting on a body exceeds a certain limit, per- 
manent distortion takes place, or the body receives a *' set "; 



lO 

but under this limit when the force is removed the sub- 
stance returns exactl}^ to the original shape. There is then 
a certain point called the Elastic Limit beyond which ma- 
terial cannot be stretched or compressed without permanent 
distortion. If the stress is continued rupture finallj' occurs. 
Both the Elastic Limit and Ultimate Strength depend upon 
the nature of the material and vary greatly. 

Let the total elongation or stretch of a rod subjected to 
a stress P, be K inches and its length 1 inches. Then the 
unit elongation or the stretch of each inch of length is 

K 
k = — , [2] 

Thus if a rod 12 feet long is stretched 0.5 inch, the unit 
stretch is k = ^ ^:^^ ^ — .0035 inch. 

Prob. 5. What unit stress would be caused b}- a weight 
of 2 tons hung on a steel bar whose weight per foot is 15 
lbs.? Ans. 902 lbs. per sqr. inch. 

Prob. 6. A stick of white pine 2 inches square and 
weighing 20 lbs. is stretched o.oi foot. What is the unit 
elongation ? Ans. 0.005 inch. 

Art. 8. Experimental Laws. 

Many experiments have been made upon the materials 
used in construction in order to establish the ultimate 
strength, elastic limit and the elongation, and from these 
the following experimental laws have been deduced. 

(A) When the Elastic Limit is not exceeded, the elon- 
gation is directly proportional to the stress. Thus if with 
a rod of a given length and section, the stress is doubled, 



11 

the elongation is also doubled; if increased three times, the 
elongation is also three times, etc. 

(B) The elongation is inversel}^ proportional to the 
cross section. Thus with a rod of a given length and sub- 
jected to a constant stress, b}^ doubling the cross section 
the elongation will be reduced one half, etc. 

(C) The elongation is directly proportional to the 
length. Thus under the same load and section, a rod 
twice as long will stretch twdce as far, etc. 

(D) Beyond the Elastic I^imit law^s A, B, and C no 
longer hold true. 

(K) Suddenly applied stresses are more injurious than 
those gradually applied. 

(F) The ultimate strength of material under repeated 
applications of a stress becomes less, depending on the fre- 
quency and range. 

(G) When the stress ranges from o to about the Elastic 
Limit, this limit must be considered'as the ultimate strength 
after a very great number of applications. 

(H) About one half the elastic limit must be considered 
as the ultimate strength, if the stress varies continually 
from compression to tension between these half limits. 

The last three laws simply effect the ultimate strength 
and should be considered in proportioning the parts of 
machinery or structures subjected to varying stresses. For 
practical purposes they may be expressed by the following 
empirical equation:* 



* See Merriman's " Mechanics of Materials.' 



12 



u = 



211 



s 



+ 



211 



;e 



S' 
S 



[3] 



Where U = ultimate strength under a varying load, 
u = ultimate strength under a stead}^ load; 
e izz unit stress at elastic limit. 

S z=z greatest unit stress the material will have to stand. 
S' = least unit stress the material will have to stand. 

Either S or S' should be considered minus if one is com- 
pressive and the other tensile. For example if S' = S or 
there is a steady load, the equation becomes U == u. On the 
other hand if S' = — S or the range of stress is from com- 
pression to equal tension, the equation gives U = -^, or the 
ultimate strength under these conditions is only one half 
the unit stress at the elastic limit, in agreement with law 
(H). vSimilarl}' when S'' == o, U = e in agreement with 
law (G). 

The following tables give the ultimate strength under a 
stead}' load, the elastic limit, and coefficient of elasticity 
of the more important of the materials of construction. It 
should be remembered that average values onh^ are given, 
the range in strength of the different grades of the same 
substance being ver}- great. In all important work the 
strength of the material should be found by actual test, or 
bought under a guarantee. 



13 

TABLE IV. 

Tablk of Ultimate Strength in pounds per 
square inch. 

Coiupressiou. Tension. Shear. 

Cast Iron 100,000 20,000 15,000 

Steel, soft 80,000 71,400 57,000 

" Plates 70,000 71,400 5 7, coo 

" Wire 130,000 

" Hard 107,000 107,000..- 85,700 

Cast Steel, soft 143,000 114,000 91,400 

" ** hard 295,000 143,000-. 

** *' wire .-. 160,000 

Wrought Iron 60,000 57,000 45)7co 

" " Plates 43,000 47,000 37,000 

" Wire 86,000 

Copper Sheet 58,300 30,400 25,000 

Gun Metal 38,000 26,800 

Brass, Common 42,000 30,500 

Phosphor Bronze 58,000 

Aluminum 9,700 

'' Bronze (Cu. 89.5) 72,000 

White Oak 10,000 11,400 2,300 

" Pine 6,000 6,600 1,860 

Yellow '' 7,300 8,000 

Beech 9,400 9,000 

Granite 8,000 

Sandstone 6,000 1,200 3,000 

Brick ] ^■'^° 

1. 5.000 



H 



TABLE V. 

Tabi^k of Unit Stress:e;s at the Ei^astic Limit and 
Coefficients of Elasticity. 

Limit of Elasticity. Coefficients 

Compression, Tension. Shear. of Elasticity. 

Cc Gt ^s ^ 

Cast Iron 21,400 10,700 8,600 14,000,000 

Steel, Soft 28,600 28,600 23,000 29,000,000 

" Plates 36,000 36,000 28,600 

' ' Wire 64,300 

" Hard 38,600 38,600 30,800 32,000,000 

Cast Steel, Soft 71,400 53,600 40,000 34,000,000 

" " Hard 95,000 76,200 

" " Wire : 43,000,000 

Wrought iron bar 20,000 20,000 16,000 28,700,000 

" " plates 20, 000 20,000 16,000 26,000,000 

'' " wire 31,400 31,000,000 

Copper Sheet 16,000 10,200,000 

Gun Metal i5>5oo 14,000,000 

Brass, Common .8,000 13,400,000 

Phosphor Bronze 19,700 14,000,000 

Aluminum 2,230 5,900,000 

" Bronze ) 28,000 6,900,000 

Cu. 89.5 Al. 10.5 [ 

White Oak 2,570 3,000 300 1,700,000 

" Pine : 1,000,000 

Yellow Pine i ,600,000 



15 

Prob. 7. Compare the tensile strength of a bar of 
wrought iron one inch square and ten feet long with that 
of a bar of aluminum of equal length and weight. 

Ans. Ratio = | ^ 

Prob. 8. If a force of io,oco lbs. causes a bar of iron to 
stretch 0.25 inch, how much would a force of 30,000 lbs. 
stretch a bar of twice the length and three times the cross 
section ? Ans. 0.5 inch. 

Prob. 9. The stress in a soft steel tie rod ranges in ten- 
sion from 80,000 lbs. to 40,000 lbs. Find the probable 
ultimate strength of the rod in lbs. per square inch. 

Ans. 46,400 lbs. per sqr. inch. 

Prob. 10. What is the ultimate strength in lbs. per 
square inch of a soft steel piston rod used in a steam 
engine ? Ans. 14,300 lbs. per sqr. inch. 

Art. 9. CoKFFiciKNT of Ki^asticity. 

The ratio of the unit stress to the unit elongation caused 
by it is called the Coefficient of Elasticity^ and since by law 
(A) the elongation is directly proportional to the stress, 
this ratio is constant. Designating it by the letter E, we 
have by definition 

.. s • 

^ - —- [4] 

k 
Values of K for various materials are given in Table V. 
Substituting values of S and k taken from equations [i] 
and [2], we have 

P K PI 

E:= = or Kzz [5] 

A 1 AE 



i6 



For example, if a wrought iron bar 2 inches square and 
100 feet long is extended by a force of 20 tons, the total 
20 X 2,000 X 100 X 12 



elongation K 



4 X 28,700,000 



0.42 inch. 



Art. 10. Elongation of a Rod by its Own Weight. 




Fig. I 



m n 


y 


B D 


1 


Hence mn 


::^ 



izi -V- or m n = 



w A 1 y 

1 



Let B C represent a rod sus- 
pended verticall3^ The stress 
due to its own weight will evi- 
dently be greatest at the point B 
and zero at the lower extremit)'. 
Designating its length in inches 
b}^ 1, its cross' section in square 
inches b}- A, and the weight of a 
cubic inch of the material b}- w 
pounds, the total stress at B is 
P =: wAl pounds. Lay off BD 
equal to this stress, join D and 
C by a straight line, and at any 
distance j draw mn parallel to 
B D. From the similar triangles 
BDy 



1 



But BD = v/Al. 



= w^ A y. But w^ A y is the weight 



of the rod below the point m, and consequently the total 
stress at this point. The average stress wnll evidently be 



BD + o 



BD 



or average P := 



wAl 



22 - 2 

Substituting this in equation ( 5 ) , the total elongation 



17 

w A ? w ? 

K = = [6] 

2AE 2 E 

Thus if the iron bar 2 inches square and loo feet long 
were suspended vertically with a load of 20 tons at the 
lower end, to the elongation of 0.42 inch due to this load 
must be added that due to the w^eight of the bar; or 

^_ 0.278 X 1,200 X 1,200 . , mt, . . 1 . 

K = = 0.007 inch. The total is 

2 X 28,700,000 

therefore 0.42 -|- 0.007 ^^ 0.427 inch. 

Prob. II. Find the elongation of bars of cast iron, soft 
cast steel, wrought iron, and white oak, each one inch 
square and 10 ft. long under a load of 3000 lbs. 

Ans. 0.0257; 0.0106; 0.0125; 0.211 inch. 

Prob. 12. The observed stretch of a wrought iron rod 
10 inches long and ^ inch diameter was 0.007 inch, the 
load being 6000 lbs. Find the coefficient of elasticity. 

Ans. E = 27,940,000. 

Prob. 13. Find the elongation of a soft steel rod 20 ft. 
long, when stretched to the elastic limit. Ans. 0.237 inch. 

Prob. 14. A vertically suspended wrought iron rod one 
inch square and 100 ft. long, sustains a load of 10,000 lbs. 
at its free end. Find the entire elongation. Ans. 0.4185 inch. 



Art. II. Thk Resii^ience of Matkriai^s. 

When material is elongated or compressed it is necessary 
to overcome molecular resistances, and internal work 



i8 

must be done. This iuternal work has been called Resili- 
ence, and it is measured b}' the average stress multiplied 
hy the distance through which it moves. Thus if a load P 
is gradually applied, the elongation increases from o to K 

P 

m a direct ratio. The average stress is — , and since it 

2 



K P 



acts through a space K, the resilience L 

2 

Substituting the value of K from equation [5], 

P'^ 1 
Iv = :7 inch pounds [7] 

2 A E 

Thus if an iron bar 2 inches square and 100 feet long is 
stretched b3' a force of 20 tons, the resilience 

40,000X40,000X1.200 o ^ • 1 J 

L = ^^ — =: 8,360 inch pounds. 

2 X 4 X 28,700,000 

If a lead be suddenh' applied, the external work will be 
the product of the entire load P bj'the elongation; the inter- 
nal resistance however, at first is zero, and hence at the end 
it must be twice as great as before in order that the external 
and internal work ma^- equal each other. Such a load 
therefore, as for example, a falling weight v/ill produce a 
stress twice as great as one gradualh' applied. 

Prob. 15. A piston of 10 inches diameter is acted upon 
by steam having an effective pressure of 100 lbs. per sqr. 
inch. What is the elongation of the soft steel piston rod, 
its diameter being \}i inches and length 30 inches ? How 
many foot pounds of work were used in causing this elonga- 
tion ? Ans. 0.0066 inch, 2.16 foot pounds. 



19 

Prob. i6. If a wrought iron tie rod whose diameter is 2 
inches and length 30 feet, were strained up to the elastic 
limit 500 times per minute, how many Horse Power would 
be used in doing the work ? Ans. 9.95 H. P. 

Art. 12. TensiIvK and ComprEvSsiye Stresses. 

When material is subjected to tension only, formula (i) 
P =: A St may be directl}^ used, St being the greatest unit 
stress, A the cross section in square inches and P the load 
in pounds. 

The formula how^ever can only be used for compression 
when the piece is comparatively short. Should the length 
exceed about ten times the leavSt transverse diameter, the 
piece or column is always subjected to a bending action 
which materially affects the strength. This case vvdll be 
considered in a later chapter. 

Art. 13, Factor of Safety. 

The allov/able or safe unit stress to which material maj^ 
be subjected depends much upon w^hether the load is steady, 
var3dng, or suddenly applied. From formula [3] we have 
seen that the ultimate strength varies greatly with these 
conditions, and in order to provide for them it is usual to 
introduce what is known as a Factor of Safety. Thus if 
the conditions are such that the piece should never be sub- 
jected to a stress greater than one fourth of the breaking 
stress under a steady load, the factor of safet}^ F = 4. 

If only one tenth the breaking stress, F := 10, etc. 

The following table gives the factors of safet}' often 
used in practioe. 



20 



TABLE VI. 
Factors of Safety. 

Dead Live IvOad. Shocks. 

IvOad. Bridges. Machines. 

Cast iron, brittle metal and alloys 4 6 to 10 10 to 15 

Wrought iron and soft steel 3 5 to 8 9 to 13 

Cast steel 3 5 to 8 10 to 15 

Copper and other soft metals 5 6 to 9 10 to 15 

Timber 6 8 to 12 14 to 18 

Stone and brick 10 15 to 20 

When however, equation [3] is used for finding the break- 
ing stress, it is usual to adopt the factor of safety for dead 
loads. Thus, taking the case of a steel piston rod subject- 
ed alternately to equal tension and compression, S' = — S, 
and the ultimate strength U = e -4- 2 = 14,000 pounds 
per square inch. Adopting the factor f zzr 3, we have a 
working strength of 14,000 -^ 3 = 4,660 lbs. per sqr. inch. 

The student should remember that the values given in 
the tables are only averages and that often the factor of 
safety mu.st be made much greater to meet peculiar require- 
ments. 

Prob. 17. What should be the cross section of a soft steel 
tie rod in a bridge, where the total stress ranges from 80,000 
lbs. to 40,000 lbs.? Ans. 7 sqr. inches. 

Prob. 18. Find the lengths of vertically suspended rods 
of wrought iron and aluminum which would break from 
their own weight. Ans. Iron 17,080 ft.. Aluminum 8,692 ft. 

Prob. 19. Find the height of a granite tower of uniform 
section which would crush tinder its own weight. 

Ans. 6981 ft. 



21 



Prob. 20. Find the elongation of a soft steel rod suspend- 
ed verticall}^ when strained to the elastic limit b}^ its own 

Ans. 50.1 inches. 



weight 



Art. 14. Shearing Strkssks. 



A force applied as shown in Fig. 2, 
produces what is known as a Shearing 
Stress; taking the name from the action 
of a pair of shears. Denoting the force 
by P, the resisting cross section by A, 



and the unit shearing stress by Ss we have Fi 




P =r A S 



[8] 



For example, Fig. 3 illustrates two plates held together 
by a rivet. The force P tends to shear it off where the 
plates join. Denoting the di- 
ameter of the rivet by d, the P p^ 
formula becomes "A*^"^"^ IrVv'V v P 

Ss 



C\ 




J. 



4 






Fig. 4 illustrates plates held together hy a bolt and nut. 
The tension P produced by turning 
the nut tends to shear off the bolt 
head along the dotted lines, to strip 
the threads, and to pull apart the bolt. 
If t is the thickness of the head, and 
d the diameter of the bolt, the 
section resisting shear is A = zrdt. Fiq. 




Hence P n: ,t d t Ss The tensile stress produced in 

/T d' St 

the bolt, from equation (i) is P =z 

4 

Hence in order that the strength of the head may equal 

that of the bolt, 



;r d t S 



^'s 



/Td^St dSt 
or t = — 

4 4Ss 



Thus if the diameter of a steel bolt is one inch, its ultimate 
tensile strength being 71,000 lbs. per sqr. inch, and its ulti- 
mate shearing strength 57,000 lbs. per sqr. inch, the bolt 

I X 71,000 



head should be t 



o.^i inch thick in order 



4 X 57 )000 

to be as strong as the bolt. In practice t is commonly 
made equal to d, and hence bolts seldom fail through 
shearing. 



S /^ 



Fig. 5 illustrates a punch and plate. 
Here again the shearing resistance of the 
plate P = ;r d t Ss 

The compressive or crushing stress in 

^^^^ , „ 'r d' Se 

the punch r =: 

Fig. 5. 4 



When these are equal t = 



da 
4S, 



Prob. 21. What force must be used to punch a ^'s inch 
hole through a % inch wrought iron boiler plate ? 

Ans. 44,860 lbs. 



23 

Prob. 22. How thick a wrought iron plate could be 
punched by a hardened cast steel punch whose diameter 
equals one inch ? Ans. 1.61 inches. 

Prob. 23. What force would shear off the head of a cop- 
per bolt, the diameter of the bolt' and the thickness of the 
head being each one inch ? • Ans. 78,550 lbs. 

Prob. 24. The diameter of a soft steel piston rod is sV^ 
inches, the diameter of the piston being 24 inches and the 
steam pressure 160 lbs. per sqr. inch. What factor of safe- 
ty does this imply? Is it safe? Ans. 2. [See Prob. 10.] 

Art. 15. Stress Produckd by Change of 
Temperature. 

When material is heated or cooled it usually expands 

or contracts. Thus let a bar whose length is 1 inches, be 

increased in length by k inches when heated from t° to ti F. 

k 
The increase of length for one degree is inches, and 

k 

for each inch of length it is r— -r inches. This quantity 

i(t I i) 

is called the coefficient of linear expa7isio7i. Designating it 

by the letter a we have, 

k 
2.— -T— or k = al(ti — t) 

l(ti — t) 

Average values of (a) for some of the more important 
materials used in construction are as follows : 
For brick and stone, a=:o. 000 0050 
For cast iron, a=o.ooooo6 2 

For wrought iron, a=::o.ooooo6 7 
For steel, a^o.0000065 

For copper, a=:o.ooooo9 7 



24 

If, however, material v/hen heated or cooled is prevented 
from expanding or contracting, the stress produced is the 
same as that which would have occurred had the piece 
been stretched or compressed by some outside force an 
amount equal to that the heat w^ould have caused had 
there been no restraint. To determine this stress we may 
make the value of K in formula (5) equal to that found 

PI SI 

above, or . .^ = -^ =z a 1 (ti — t) or 

S = a (tx-t) E (9) 

Thus let an iron bar 10 feet long be heated from 50° to 
500°. The total increase in length, if free to expand, is 
K = 0.000006 7 X 10 X 12 X (500 — 50) == 0.56 inch; but if 
this increase is prevented, the unit stress caused by the heat 
is 8 = 0.0000067 X(5oo — 50) X 28,700,000 = 86,500 lbs. 
per sqr. inch. This being greater than the ultimate strength 
of iron in compression, the bar w^ould be distorted. 

Prob. 25. A wheel tire of wrought iron w^hen heated 
from 70 to 600 degrees is to slip over a wheel whose diameter 
is 4 feet. What should be the length of the tire when 
cold? Ans. 12 ft. 6j{ inches. 

Prob. 26. What unit stress would be caused were the tire 
prevented from contracting more than one half the amount 
it had expanded? Ans. 50,956 lbs. per sqr. inch. 

Prob. 27. A continuous railroad track, with welded joints 
was laid when the temperature of the rail was 60° F. As- 
suming it to be held securely by the spikes, what unit stress 
w^ould be caused in the iron; first, when the temperature fell 
to — 30 degrees? second, when it rose to -{- 120 degrees? 
Ans. St = 17,306 Sc = ii>537 lbs. per sqr. inch. 



CHAPTER III. 
STRESSES IN, AND STRENGTH OF BEETING. 
Art. i6. Fundamkntai, Equations. 

The universal use of belting for the transmission of 
power makes a knowledge of the principles involved of 
especial importance. 

Let Fig. 6 represent a pulley 
over which a flexible band passes, 
and for the present let us consider 
the pulley as fixed. Eet Ti repre- 
sent a force applied on one side, 
and T2 another force on the oppo- 
site great enough to cause the belt Fig. 6. 
to slip on the pulley; and let us determine the relation 
between these two forces, the angle q) on the pulley em- 
braced by the belt, and the coefficient of friction. 

As in many of the most important deductions in applied 
mechanics, it seems necessary in this instance to use sim- 
ple calculus. The student who is not familiar with the 
elements of this branch of mathematics must be content to 
accept the results on faith, or better, should study and 
make himself familiar with a mathematical tool of great 
practical importance. 




26 




In Fig. 7 let ab = ds repre- 
sent a small arc of a pulle}^ 
over wliich a belt passes, 
r = radius of the pulley, 
p = normal pressure the belt 
exerts upon the pulley in 
pounds per unit of length. 
T = the tension in the belt 
at a. 

dT = the slightly increased 
tension at b due to friction of 
the belt on pulley, 
u = coefficient of friction. 

Since the normal pressure 

per unit of length is p pounds, 

pounds, and in the length ds 

to upds pounds; that is to say, the 

dT = upds. But ds = rdq). 

Hence dT = uprd fP (a) 

We have then a force of T pounds acting tangentially 
at a, and if we should appl}- another tangential force of 
T + dT pounds at b, the belt would be just on the point 
of slipping. 

Let us now find an expression for the unknown normal 
pressure p . In Fig. 8 draw" de and ef parallel respec- 
tively to ac and be, and from the point of intersection e 
lay off ed = T , and ef = T -f dT . 

These forces being in equilibrium, we may find their 
resultant by completing the parallelogram defg . Calling 
the resultant R and noting that the angle edg = dmf = 

dcp 



Fig. 8. 
it causes a friction of u p 
this would amount 
increase in tension 



aob = dg^ , we see that 



2 



= T sin- 



but since the 



27 

angle is very small, the sine is equal to tlie arc itself, or 
sm— ^ = -—^ Hence — = ^— or R = Td^p 

2 2 2 2 

But the resultant pressure R on the arc ds must also 
be equal to the unit normal pressure p multiplied by the 
length of this arc, or R = p ds . Hence p ds = l^d^^j 

or p = — ; ; but ds = r d^ . Therefore the norm-al 

ds 

Td^ T ,, . 

pressure p = = (b) 

rdcp ^ 

Substituting the value of p in equation (a), 

T dT 

dT = u r d(p = uTdcp or u d^ = -7—- 

r 1 

Integrating this between the limits T2 and Ti , 

-^ = nat. log. -^ [10] 

This is the desired equation showing the relation between 
the tension on the two sides of a belt, the angle in circular 
measure embraced on the pulley, and the coefncient of fric- 
tion. We may avoid the use of natural or hj^pobolic loga- 
rithms by multiplying the second member by 2.3026. The 

angle ^ is equal to — - — . Hence our formula becomes 
"78^ = 2.3026 log^^, or 

0.00758 u^° = log^ [11] 

which is in more convenient form for practical use. It 
will be observed that theoretically the friction is independent 
of the diameter of the pulley, although it is probable that 
the stiffness or rigidity of a belt or rope is a material factor 



23 

when the diameter is small. The principle error however 
arises in uncertainty as to the proper coefficient of friction 
of the belt or rope upon the pulle3^ For leather belts on 
iron pulleys under ordinary conditions of working u varies 
from 0.3 to 0.4. Experiments b}^ Briggs and Towne show 
that the latter value may safely be taken. For iron rope 
on an iron pulley u is about 0.15, while if the bottom is 
lined with leather or rubber it may be from 0.25 to 0.3. 

The following table, giving the values of ~ for angles 

J- 1 

varying from 30 to 300 degees and for a coefficient of friction 
varying from 0.2 to 0.5, has been calculated by the use of 
equation (11) and will include all cases likely to occur in 
practice. 

TABLE VII. 

T2 
Values of 7=^ = r 



T 



Angles embraced bj- belt , ^ „ 

° . . ^ u = o.2 u = 0.3 u = 0.4 u = o.5 

m degrees. xj -r yj 

30 I. II 1. 170 1-233 1-297 

45 1. 17 1.266 1-369 1. 481 

60 1-233 1-369 1 -52 1 1.689 

75 1-299 1. 48 1 1.689 1-924 

90 1 -369 1.602 1-874 2.193 

105 1-443 1-733 2.082 2.500 

120 1 -52 1 1-875 2.312 2.851 

135 1-602 2.027 2.565 3.247 

150 1.689 2.194 2.849 3-702 

165 1778 2.372 3.163 4.219 

180 1.S75 2.566 3.514 4.808 

195 1-975 2.776 3.901 5.483 

210 2.082 3.003 4.333 6.252 

240 2. 31 1 3.514 5.340 8. 119 

270 2.566 4. 1 12 6.589 10.550 

300 2.S49 4.808 8. 1 17 13-700 



29 

As an example of the use of the table, let vis suppose a 
weight of ICO lbs. to be fastened to a rope passing over a fixed 
pulley. What force applied at the other end would lift the 
weight, the angle (p being 150 degrees and the coeincient 
of friction being 0.3 ? From the table, 

T 

— ^ = 2.194 or T2 = 100 X 2.194 ^^ 219.4 pounds. 

If the angle had been (p = 270°, or three fourths a turn 

T 
-7=^ = 4. 112 or T^ = 411. 2 pounds. 

-L I 

If the rope had made a complete turn, or ^ = 360°, 

T 
from the formula, log-7=r- = 0.00758X0.3X360 =: 0.81864 

Ii 

T 
Hence -=^ = 6.56 and T^ = 656 pounds. 
1 1 

If the rope makes two complete turns, or ^ = 2X360, 

T T 

log ^^ = 1.63727 -rf- = 43-36 and T, = 4,336 lbs. 

or, conversely, a man exerting a force of 100 pounds might 
with two turns of his rope about a post, hold a weight of 
over 2,000 pounds. 

Prob. 28. A weight of 400 lbs. is attached to a rope 
passing over a fixed pulley. What weight fastened at the 
other end would cause the rope to slip when u =: 0.35 and 
(p = 180° Ans. 1201.2 lbs. 

Prob. 29. A force of 219.4 ^^s. applied to a belt causes 
it to slip on a fixed pulley and lift a weight of 100 lbs. 
What was the coefficient of friction, the angle cp being 150°? 

Ans. u = 0.3. 




Fig. 9. 



30 



Prob. 30. In Fig. 9 let the 
two weights W, = 500 lbs. and 
W^ rest against surfaces in- 
clined 60°. Let the angle <p = 
120°, the coeff. of friction u = 
0.4 and of the weights on the 
surfaces u, =0.3. How heavy 
must Wg be in order to pull up 
W , , the pulley being fixed ? 

Ans. 1639.6 lbs. 



Prob. 31. In the last problem assuming the pulley free 
to turn without friction, how heavy must W2 be ? 

Ans. 709.5 lbs. 



Art. 



Transmission of Power. 



Fig. 10 represents two pulleys driven by a belt m^oving 
in the direction of the arrows, the pulley AB being the 
driver, and CD the follower. 

Let T^ = tension in the 

tight side of the belt. 
T, = tension in the slack 

side of the belt. 
P = force which must be 
applied at the circum- 
ference of the follower 
in order to turn it. 
V = velocity of the belt in 
feet per second. 




31 

D = diameter of the driver in inches. 
d = diameter of the follower in inches. 
N = revolutions per minute of the driver, 
n = revolutions per minute of the follower, 
u = coefficient of friction. 
H = horse power transmitted. 

The tangential force of P pounds applied to the pulley 
CD would move in one second through a distance of V 
feet. The work done is therefore P V foot-pounds per 
second; or, since a horse power is work at the rate of 550 

P V 

foot-pounds per second, H =^ -^ . But the force of P 

pounds is due to the difference in the stress in the belt on 

the tight and loose sides, or P = Tg — T, Substituting 

V (T — T ) 

this, H = — ^^^-^ ^^^ 

550 

The velocity V may be expressed more conveniently in 
terms of the revolutions per minute of either pulley and its 
diameter. Thus, in one turn, a point on the circumference 

TTD 

of AB will move through a distance of feet, and in 

TT D N 

one second through a distance of — feet, or 

^ 12X60 

^D N TTd n 

V = , or considering the other pulley, V = 

720 ir J ^20 

Substituting above and combining the constants into one, 
^ ^ D N (T, - T, ) ^ dnCT, - T, ) 

126,000 126,000 

or, solving for T^ — T, , 

^ ^ 126,000 H -, 

^^ ~ ^^ "^ DN ^'^-' 



32 

The ratio of T^ to T, may be found from formula [ii] 

T 
or from the table, and calling this ratio r, --— = r, or 

-*■ 1 

T2 

T, = — ~ Substituting in [12] and solving for T2 

^ 126,000 H 



DN 



X;^ [13] 



When the pulleys are at rest, or when the belt is first 
put on, the stress T in the two sides is the same. When 
started, the stress on one side increases to Tg and is redu- 
ced to Ti on the other, but the average stress remains 
nearl}^ constant. Therefore we maj- write another equation 

T + T 
^ = ^ \ ' [14] 

With the aid of these equations, we shall be able to find 
the necessary initial stress T in a belt, and the after stresses 
Tj. and Ti on each side, w4ien a given povv^er is to be trans- 
mitted, and to solve most of the other problems arising in 
practice. Thus, let it be required to transmit 10 horse 
power from one pullc}^ to another, the smaller pullej^ to 
make 200 revolutions per minute and to be 2 feet in diam- 
eter. The angle embraced by the belt wall evidentl}'- 
depend upon the diameter of the larger pullej^ and upon 
its distance ; but let us assume ^ = 165 degrees, and 

T 
u ■=. 0.4. From the table, -^;^ zi= r = 3.16, and from 

^ I 

, N m 126,000X10X3-16 ^ , 

equation (13) T, = ,4X200X2.16 ^ ^04 pounds. 
Since this is the greatest stress the belt will be called 
upon to bear we can, as explained in the next article, fix 
upon the size of the belt necessary. 



33 

^ . / N MA ^rA 126,000 X 10 

From equation (12) T^ — Ti = — = 262.5 

^ \ 2 24 X 200 

pounds, and T, = T^. — 262.5 = 121. 5 pounds. 

Finall}^ in order to create sufficient friction, the tension 

or stress when the pulleys are not running should be 

T ^ T 

T =z — - — ^ =z 323 pounds. 

Prob. 32. Two pulleys have each a diameter of 40 
inches, and make 120 r.p.m. and the belt is to transmit 
10 horse power. What would be the stresses T, T, and 
T^ when u =: 0.4? *Ans. 235.4; 104; 366.8 pounds. 

Prob. 33. A belt connects a fly-wheel of 20 feet diameter 
with a pulley of 8 feet diameter, the distance from center to 
center being 20 feet. Required the tensions in the belt 
when transmitting 70 horse power with 70 R. p. m. of the 
fly-w^heel, when u = 0.4? Ans. T, = 324, Tg = 924 lbs. 

Prob. 34. A belt transmitting 10 horse power connects 
pulleys of equal diameter. The difference between T^ and 
T, is 262 pounds and u = 0.4. What is the diameter of the 
pulley? Ans. 40 inches. 

Prob. 35. The tension in a belt connecting two equal 
pulleys is found by experiment to be 235.4 pounds when at 
rest, and also that a force applied at the circumference of 
262 pounds would cause the belt to slip. What was the 
coefficient of friction ? Ans. u = 0.4. 

Art. 18. Thk Strkngth of Lkather Bklting. 

The ultimate strength of leather used for belting varies 
from 3,000 to 5,000 pounds per square inch, but when, as 
usual, laced joints are used this is reduced from 900 to 1,500 



34 

pounds per square inch. It is usual to adopt a safe working 
strengtli of about 320 pounds per square inch. Formula ( i ) 
therefore applied to leather becomes 

T2 =: S A r= 320 A, or T2 = 320 t b, 
when t equals the thickness of the belt and b its width in 
inches. The following table will be of service in using 
this equation, the last row being the working strength 
per inch of width. 



TABLE VIII. 

t=-3_J7_l._5_ 3. J7_ i 5 

1G32 4 IG 8 16 2 16 8 

320 t = 60 70 80 100 120 140 160 180 200 

The ordinary thickness of a "single" belt is -^^ and of a 
''double" belt yq of an inch. Thus, in the problem worked 
out in the last article, Tg = 384 pounds, and if we assume 
a " single " belt, the width should be 

b = — 384_ ^ _38^ ^ i^^^^,3^ 

320 X t 70 -^ ^ 

Prob. 36. What .should be the width of the belt in Prob. 

7 ' 7 • • 

-12 if -^ inch thick ? What in Prob. 32 if -V inch thick ? 

7 
Prob. 37. A belt 5 inches wide and -^ inch thick passes 

3^ 

over a 10 inch pulley making 500 R. p. m. Assuming the 
stress not to exceed 70 pounds per inch of Vsddth, that u = 
0.4 and ^ = 165, what horse power could the belt transmit ? 

Ans. 9.5. 



35 



Art. 19. RoPK Belting. 

The formulae deduced in the preceding articles appl}^ also 
to rope belting, although modified somewhat by the use of V 
pulle3^s which materially increase the friction. Let Fig. 11 
represent a section of rope and V 
pulley, the sides of which make an 
angle a with each other. Let p 
be the normal pressure which the 
tension in the belt would create 
upon a flat pulley, and make the 
line OA equal to p. Resolve this 
force into components at right 
angles to the sides as shown. 
Then O B is the normal pressure pi acting upon each side. 

P . OL 

— ~ — '- sm — 
2 2 




Fig. 



II. 



From the figure pi 



The total normal 



pressure is 2 p] 



a 



sin — and the friction, 
2 



2 upx 



a 



u p -f- sin — But u p is the friction upon a fiat pulley. 

Hence we see that the friction upon a V pulley is equal to 
that upon a fiat pulley divided by the sine of one half the 
angle made by the sides with each other. Letting u -^ 

OL 

sin — = Ui , formula (11) becomes when applied to a rope 
and V pulley 

.007 58 Ui cp 



log 



T. 



T. 



[15] 



The following table gives values of Ui , corresponding to 
various values of the angle a and u. 



z^ 



TABLE IX. 

Angles of sides Values of u, , when u = 

= ^° 0-I5 0-2 0.25 0.3 0.35 

30 0.5S 0.77 0.97 1. 16 1.35 

35 0.50 0.66 0.83 I. GO I. 16 

40 0-44 0-58 0.73 0.S8 1.02 

45 0-39 0.52 0.65 0.78 0.91 

The usual angle for hemp or cotton rope is 45 degrees. 
For greased wheels with this angle u, :=: 0.4, while for 
ropes and pulleys in ordinary working order u, may be 
taken at 0.7. For steel and iron cables it is usual to make 
the bottom of the groove wide enough to allow the cable 
to rest on it and the friction is not materially increased 
over that on a fiat pulley. 

For example let it be required to transmit . 50 horse 
power by means of rope belting, the diameter of the 
driving pulley being 3 feet and its speed 100 revolutions 
per minute. Let u, = 0.7 and q) = 180 degrees. Sub- 

T^ T^ 

stitutmg m formula (15), log — ^ = 0.955, -=- = 9.01 

'^ I 1 

^ r I / X ATA ATA 126,000 X 50 

From formula (12) T, — T = ' ^ ^ = 1,750 

36 X 100 

pounds, and from formula (13) 

T, = 1,750 X-|^ = 1,970 pounds. 

Hence T, = 1,970 — 1,750 = 220 pounds, and the initial 

^ i,Q7o -I- 220 J 

stress T = -^-^ = 1,095 pounds. 

2 

The greatest stress in the rope is therefore 1,970 pounds. 



37 

Prob. 38. In the example just explained, wliat must be 
the value of u, and of the angle oc in order that the stress 
T, may equal 100 pounds ? Ans. u, ::= 13.5, 3°, 21'. 

Art. 20. Tkk Strength of Ropes. 

The material usually employed is hemp or cotton. The 
ropes vary in size from i to 2 inches, and the net section is 
about 0.9 of the area of a circumscribed circle. The ulti- 
mate strength varies from as low as 7,000 pounds per square 
inch to as high as 1 2 ,000 pounds per square inch . According 
to Low & Be vis,* 1,200 pounds per square inch is considered 
a fair working stress, but for great durabilitj' the general 
practice is not to allow more than 140 pounds per square 
inch of actual section, giving a factor of safety of about 60. 
The weight of a rope when dry is approximately given b}^ 
the formula W = o.3D^ Vvdiere W is the weight in pounds 
per foot, and D the diameter of the rope in inches. 

Let us now^ find the proper number of i^ inch ropes to 
be used in the example of the preceding article. The 
cross section corresponding to i^ inches is 1.23 sqr. inches; 
the effective area is 1.23 X 0.9 = i.i sqr. inches. The 
total stress in each rope therefore would be i.i X 140 = 
154 pounds, and dividing T^ = i>97o by this, we find 12 
ropes are necessary. 

Prob. 39. An engine developes 500 h. p. at a speed of 
120 R. p. M. The fiy-wheel is 15 feet in diameter, from 
which the power is taken by means of cotton ropes. As- 
suming the diameter of each rope to be ij^ inches, how 
many are needed to transmit the power, the angle (p 
being 150° and Uj = 0.7 ? 

Ans. T^ = 3472 pounds; 15 strands needed. 



# Manual of Machine Design. 



38 



Art. 21. Influkxch of Ckntrifugal Force. 



When a pulley runs at a liigh velocity the influence of 
the centrifugal force upon the belt is marked and should be 
considered in finding the necessary tension and friction. 
This force tends to diminish the normal pressure between 
belt and pulle}^ and hence to dimin- 
ish the friction. An increased ten- 
sion must therefore be given to the 
belt in order to counteract this loss. 
To find the increase, let figure 12 
represent a pulle}^ with a belt em- 
bracing an angle of (p degrees. 
Bisect the angle b}^ the line Oc and 
let us first find the centrifugal force 
of the belt acb in the direction of 
this line. Let the weight of the belt 
per unit of length be W pounds, and Fig. 12. 

consider a short length ds whose angular distance from 
the central line is /3 degrees. The centrifugal force dL 

W V^ ds 
acting radiall}^ of this short length is dly = 

where V is the velocity of the belt in feet per second, R 
the radius of the pulley in feet, and g the acceleration due 
to gravit}^ = 32.2 feet per second. But d s := R d^. 




Hence dL := 



W V^ d/? 



Resolving this into components 



perpendicular and parallel to the central line, and calling 

W V^ cos/? df3 

the latter dC, dC = dLcos/? = 



39 

Integrating this between the limits f3 -=^ o and /? = — 
we find for the centrifugal force in the direction oc of the 

portion cb, C = sm — ; and for both portions, or 

2 W V^ ^ 

acb, C = —sin — . Lay off this force as shown m 

g 2 

the upper part of the figure, and resolve it into two forces, 
one parallel to the line aTs and the other parallel to bTi 

Calling these Ts and Ti from the figure we see that 
t; := T: = -^ -, sin^ or T: = ^^^ [x6] 

2 2 g 

The total stress in the tight part of the belt should there- 
fore be increased by this amount, which it will be noted, 
does not depend upon the length of arc embraced on the 
pulley. The total stress in a belt is therefore, 

To = T3 + t: [17] 

The weight of leather belting is about W = 0.43 t b. 

/^i - . . , , ,^ ;r N D 

The velocity in feet per second V = - — — — -, and ^ =^ 
^ ^ 12 X 60 ^ 

32.2 feet per second. Inserting these values and reducing, 

the formula becomes 

T^ — 0.000000254 N" Dn b [16'] 

Thus in the example at the end of Arts. 17 and 18, 
N = 200, D = 24, t = -3^, and b = 5.5. Substituting 
these and reducing, T2 = 3.5 pounds, and the greatest 
stress in the belt. To = 3.5 + 384 1= 387.5 pounds. 

The speed is so slow in this case that the influence of the 
centrifugal force is small, but many pulleys run at speeds 



40 

as great as 4,000 revolutions per minute. For illustration, 
let us assume this speed, keeping other dimensions the 
same. The value of T2 would then be about 2,820 pounds, 
or much greater than the belt could bear. 

Prob. 40. The 6 inch pulley on a spindle carrying a 
buzz-saw is to make 2,500 r.p.m. and the belt must trans- 
mit 10 h.p. Find the greatest stress in the belt, and its 
width; its thickness being -^, u= 0.2 and (p = 150°. 

Leaving out centrifugal force, T2 = 205 pounds, and 
b = 3.5 inches. 

Considering centrifugal force. To = 44 -f 205 = 249 
pounds, and b should equal 4 inches. 



CHAPTER IV. 



CYLINDERS, RIVETED JOINTS, BOILERS. 

Art. 22. FuNDAMENTAi. Equations. 

Let Fig. 13 represent a cross section of a pipe or cylin- 
der whose length is 1 inches, and diameter is D inches, 
and let the pressure inside, acting 
in every direction exceed the ex- 
ternal pressure by p pounds per 
square inch. Draw an}^ diameter 
as DD, and let us find the total 
resultant pressure P acting at 
right angles to this diameter. 

Assume any narrow strip whose 
width is s I . Its area is 1 s i , and 
the total pressure acting on it is 
p 1 Si pounds. Let P' drawn ra- 
dially represent this pressure. 
Resolving it into components, one 
of which Pi shall be perpendicular, and the other parallel 
to DD, and calling p the angle between P' and Pi we 
have Pi = P' cosy6*, or Pi = p 1 Si cos/J. But from the 
Fig. Si cos/? = ai, where ai is the projection of Si upon 
DD. Hence Pi =2 p 1 ai Taking the next narrow strip, 
in a similar manner we have P2 = p 1 aa ; P3 == p 1 a3 ; etc. 




Fig. 13. 



and Pi -j- P, 
— P. But ai 



etc. =pl(ai -|- aa-r ^3+ etc.) 
a3 -f etc. = D, hence P = p 1 D. 



42 

This resultant pressure is evidently resisted by the ma- 
terial at the points D D; and calling the thickness of the 
C3-linder t, and the unit tensile stress St , the total resisting 
stress b}' equation (i) is P = 2 1 1 St Equating this to 
the pressure which causes it, we have, 

p 1 D = 2 t 1 St or p D = 2 t St [18] 

This is the fundamental equation for the strength of pipes 
and cylinders against longitudinal rupture. 

The total pressure acting upon the end of a cylinder as 
in Fig. 14 is equal to the unit pressure p, multiplied by its 

TT D^ p 

area, or P = ^-— This will be re- 

4 

sisted by a ring of metal whose thickness 
is t inches and whose area nearly equals 
^ D t sqr. inches. Hence again placing 
Fig. 14. the resistance of the material equal to 

the pressure which causes it, we have, 

TT D^ p = 4 ^ D t St or pD = 4 t St [19] 

This is the equation for the strength of cylinders against 
transverse rupture. From these equations it will be seen 
that the strength of a cylinder against transverse rupture 
is twice as great as for longitudinal. Accordingly rupture 
usually occurs by " splitting " lengthwise, although from 
flaw^s in the metal it sometimes is transverse. 

The values of St are to be taken from Table IV and a 
proper factor of safety must be chosen to ensure against 
unexpected pressures, and bad material or workmanship. 
In choosing these it should be borne in mind that wrought 
iron pipes or cylinders are usually either welded or riveted. 




In the former case the weld is geuerall}^ not as strong, ^r 
may not he as strong as the plate. The latter case will be 
considered under the head of riveted joints. 

Standard wrought iron pipe whose 
nominal diameter is i54^ inches or less, 
is " butt " welded, Fig. 15. 

Other sizes are " lap " welded, Fig. 16; 
the lap weld naturally being the stronger. 

I^arge cylinders or boilers are also 
sometimes welded. When such joints 
are well made thc}^ are found to be from 
75 to 85 per cent, as strong as the plate. 

Prob. 41. What pressure would burst 
standard iron pipes, the nominal diam- 
eters being ^, i, and 2 inches, assum- 
ing the weld to be as strong as the 
material elsewhere ? 




Fig. 16. 



Ans. 20,040; 14,550; 4,480 lbs. per sqr. inch. 

Prob. 42. In the last example, assuming the pipe to be 
"extra strong," what is the bursting pressure in each 
case ? 



Prob. 43. Prove that the formula 
holds true for thin spheres. 



p D = 4t St also 



Art. 



Pressure Caused by I^iouids. 



When the pressure is caused by a gas, as steam, weight 
need not be considered, but when caused by water or other 
liquids, it may be much less at the top than at the bottom 
of a C3dinder, as for example in a stand pipe.' 



44 

Let li = vertical height in feet of the water above the 
point considered; w -- vv^eight of a cubic foot of water: 
and p = pressure in pounds per square inch caused bj' it. 

Then the pressure on each square foot will be wh, and 

on each square men p = pounds. For fresh vv'ater 

144 

w := 62.5 pounds nearl}", hence 

p = 0.434 h [20] 

or inversel}^ h = 2.304 p [21] 

Thus a column of water a mile high would cause the 
great pressure of 0.434 X 5,280 = 2,291 pounds per sqr. 
inch at the bottom. 

Prob. 44. What head of water would burst a standard 
iron pipe whose nominal diameter is 10 inches, the strength 
of the weld being 75 per cent, of the plate ? 

Ans. 7,196 feet. 

Prob. 45. What are the constants in equations (20) and 
(21) for sea v^-ater ? Ans. 0.445 ^^<^^ 2.245. 

Prob. 46. An iron pipe, whose diameter is 8 inches and 
thickness j^ of an inch, is subjected to a pressure of 100 
lbs. per sqr. inch, \yhat will be its increase in diameter 
due to this pressure? (See formula 5.) 

Ans. 0.000492 inch. 



Art. 24. Thick Cylinders, Flues. 

When the w^alls of a C3dinder are thick compared to the 
inside diameter, formula (18) cannot be used, the reason 



45 

being that the .stress is not uniforml)^ distributed, those 
particles near the inside being strained to a greater extent 
than those on the outside. A discussion of this case is 
somewhat complex, involving a knowledge of higher math- 
ematics. The result however maj^ be expressed by the 
equation 

2 t St r -|,v. 

p m 22 

For example, the bursting pressure of a cast iron pipe lo 

inches in diameter and one inch thick by equation (i8) is 

2 X I X 20,000 ,. ...... 

p = = 4, GOG ibs. per sqr. mcli; while bv 

lo ± ± ^ 

, . .^ . 2 X I X 20,000 

equation (22) it is — =■ 3,333 10s. per sqr. 

inch; or about 17 per cent. less. 

Where flues and C5diriders are subjected to an external 
pressure, the strength is often much less than w^hen the 
pressure is inside, for the reason that the circular section is 
likely to become oval, thus increasing the diameter in one 
direction. The strength is also effected by the length, 
being nearly inversely proportional to it. From numerous 
experiments, Fairbairn deduced the following empirical 
equation for the collapsing pressure of plane boiler flues 
of wrought iron 

806,300 t^-^9 
P^ LD '^^] 

where L is the length of the flue in feet. For the working 
pressure Low & Bevis propose the equation, 

46552 t^-^9 



D 1 L 



[24] 



^Merriman's "Mechanics of Materials,"' 2nd F,d., p. 28. 



46 

These formulae however, .should onl}^ be used for lengths 
such as commonly occur in boilers. 

The strength of flues may be greatly increased by longi- 
tudinal corrugations, or by stiffening rings at short dis- 
tances apart. 



TABLE X, 



f — jL _5_ 3. _7_ 1 _JL 

^ — 4 16 8 16 J 16 

|-2- 19 —; 0.048 0.078 O.II7 0.164 0.219 0.284 



Prob. 47. What pressure would collapse an iron flue 
whose length is 10 feet, diameter 10 inches, and thickness 
% of an inch ? What is the greatest safe working 
pressure ? 

Ans. 387 lbs. per sqr. inch; 70.7 lbs. per sqr. inch, 

Prob. 48. What should be the thickness of a cast steel 
cannon capable of withstanding a pressure of 20,000 lbs. 
per sqr. inch with a factor of safety of 3, the internal diame- 
ter .being 3 inches ? Ans. 1.7 inches. 

Prob. 49. What should be the thickness of an iron flue 
to safely stand a pressure of 150 lbs. per sqr. inch, the di- 
ameter being 8 inches and the length 10 feet ? Ans. ^ in. 



Art. 25. Riveted Joints. 

Fig, 17 illustrates a lap joint with a single row of rivets 
or single riveted. Such a joint is often strengthened to a 



47 





considerable extent by 
the friction of the plates 
on each other, but since 
this is an uncertain 
quantity, it is usually 
neglected. 

I^et the thickness of 
the plate be denoted by 
t, the diameter of the 
rivets by d, and the 
distance from center to 
center in any row, or 
the " pitch," by a. Fig. 17. 

Such a joint may fail in four ways: ( i) by tearing the plate 
apart between rivets, as at n n; (2) b3'' shearing off the rivets 
where the plates join, as at s; (3) by crushing the material 
in front of the rivet, or the rivet itself; (4) by breaking the 
plate between the edge and rivet holes, as at mm. Ex- 
periment has shown that the last may readily be provided 
against by making the distance from the center of the rivet 
to the edge of the plate equal to 1.5 times the diameter of 
the rivet. 

Let P be the force acting on the length a, tending to 
tear apart the joint, and St the unit tensile stress in the 
plate. The resisting cross section of the plate is evidently 
less than before the hole was made by an amount dt square 
inches. Hence the resisting area A==:t(a — d), and 
the total resisting stress is t ( a — d ) St. This must be 
equal to the force causing it. Hence for tearing, 

P :=. t(a-d)St [25] 

In the length a, one rivet is subjected to a shearing 

stress. Its cross section is square inches, and if Ss is 

4 



48 



the unit stress we have for shearing, 
7t d^ Ss 



P = 



4 



[26] 



In the length a, one rivet is also subjected to crushing. 
The surface exposed is usually considered equal to dt, 
and if Sc is the unit stress, we have for crushing, 

P =: dtSc [27] 

If the joint is to be equally strong for tearing, shearing, 
and crushing, these equations must equal each other, or 

TTd^ Ss 



t ( a — d ) St — 



dtSc 



A joint thus proportioned would be as likely to fail one 
wa}^ as another. 

Fig. 18 illustrates a 
lap joint with two par- 
allel rows of rivets, or 
" double riveted." 

The tearing stress is 
obviously the same as 
before, but in the dis- 
tance a there are now 
two rivets subject to 
shear, and also two 
subject to crushing ac- 
tion. Hence equations 

■= 2dtSc [28] 




Fig. 18. 
[25], [26], and [27], become, 

, X r. TT d= S, 

p = t(a — d)St=2 

4 

In order that the plate may not tear along the line w in 
the Fig., Prof. Kennedy states that the net section of the 
plate, measured zigzag, should be one third greater than 
that measured along a row, or 



49 



2t(a' — d)=t(a — d) + 



t ( a — d ) 



or the diagonal pitch a' = 

Fig. 19 illustrates a butt 
joint, single riveted, and 
with double cover plates. 
The tensile strength of 
the plate is still the same. 
Each rivet resists shear 
in two places, while the 
crushing stress is borne 
by one rivet only, as in 
the first case. Hence 



P =: t(a 

2 TT d^ Ss 



• d ) St = 

= dtSc [29] 

4 

Sometimes one cover only 
is used, in which case we 
have two lap joints single 
riveted, and the stresses 
Vxdll be the same as in the 
first case. 

Fig. 20 illustrates a 
butt joint, double riveted. 
The tensile strength is the 
same, but in the length a 
there are two rivets, each 
resisting shear in two 
places, and each resisting 
the crushing force. Hence 



2 a H- d 




Fig. 19. 




Q Q 



o ' 9 

1^- a -T 

6" "6 



Fig. 20. 



50 

P 1= t(a — d) St =r "^ "^ ^' ^^ = 2dtSc [30] 

111 a similar manner expressions for the strength of 
triple and quadruple riveted joints may be found. 

The preceding equations afford a means of determining 
the proportions of riveted joints excepting when the rivets 
are subjected to twisting or bending action; or, as in boilers, 
where the joint must be calked in order to make it water 
or steam tight. The equations however are subject to 
the following practical restrictions: 

Rivets should not be placed nearer than three diameters 
or further apart than six diameters. When nearer together 
the plate is liable to damage by punching the holes in 
riveting, and when further apart there is a tendency to 
buckling when the joint is in compression. Punching is 
injurious to the plate, especiall}^ if made of steel, and the 
practice of drilling holes is rapidl}^ becoming general. 

Prob. 50. Two iron girder plates each % an inch thick 
are to be riveted together wath butt joint, double cover 
plates, and double rows of iron rivets. Find the proper 
pitch for equal tearing and shearing strength, the diameter 
of the rivets beinsr 44- of an inch. Ans. 3-J-- inches. 

" 1 o lb 



Art. 26. Efficiency of Riveted Joints. 

The ratio of the highest allowable stress in a riveted joint, 
to the highest allowable stress in the unpunched plate is 
called f/ie efficiency of the johit. Thus if we consider a 
single riveted lap joint having n rivets in a row, the 
efficiencv is 



5^ 



For Tearing et 
For Shearing eg 
For Crusliino: e. 



n t (a — d) vSt 



n a t St 


a 


n 7t d^ Ss 


TTd^ Ss 


4 n a t St 


~ 4 a t St 


n t d Sc 


__ dSc 



a St 



[31] 



n a t St 

The numerator in each case is the total stress acting 
on n rivets, or on the entire row whose length is 1 inches. 
Designating this by P and remembering that n a = 1, we 
have 

P == natSte = 1 1 St e [32] 

If we call the ultimate tensile strength of the material 



St and introduce a factor of safety f, St ^=- 



St 



Substi- 



tuting this in the above, w^e have 
1 1 St e 



P — 



f 



[33] 



where e is the least efficiency derived from equation [31]. 

Thus to find the breaking load on a riveted wrought 
iron plate 6 inches wide and half an inch thick, the 
efficiency being 60 per cent.; from equation [32] 

p =z 6 X >4 X 47,000 X 0.6 = 84,600 lbs.; 
or, if the above plate were loaded with 40,000 lbs., the 
factor of safet}- would be 

6 X >4 X 47,000 X 0.6 



f 



2.1 



40,000 
In bridge and girder w^ork, while there is considerable 
variation in practice, the following relation between diam- 
eter of rivet, and thickness of plate gives good results 
d == ii< t 



-S^ inches, 

1 o 



52 

But, owing to risk of fracture and injury to the material 
in punching holes, the diameter should not be less than 
the thickness of plate. 

Prob. 51. In Prob. 50 what is the eflficiency of the joint, 
and what is the breaking load per inch of width ? 

Ans. 77.6 per cent.; 22,110 lbs. 



Art. 27. Boiler Shklls. 



In boiler shells the pitch of the rivets is usually made 
less, especially when the plates are thin, than the equa- 
tions of equal strength call for; the reason being that 

after the plates are riveted 
the joint must be made steam 
tight by calking as shown 
in Fig. 21. The tendency 
of the calking tool is to force 
the plates apart between 
rivets, and thus the pitch is 
determined b}^ stiffness of 
the plate rather than equal tearing, shearing and crushing 
strength. The following tables taken from Low & Bevis* 
give proportions used in good practice. The efficiencies 
et and es are calculated on the assumption that for iron plates 

and iron rivets ^— = i, and that for steel plates and steel 




Fig. 



21. 



rivets 



St 



= 0.8 



* A Mauual of Machine Drawing and Design. 



33 



TABLE XI. 

Dimensions of Singi^k Riveted Lap Joints. 

Iron Plates and Iron Rivets. Steel Plates aud Steel Rivets. 



a 



et 



a 



16 
3. 

8 

7 

16 

JL 

2 

9 
1 6 

5. 

8 

1 1 
1 6 



5. 
8 
B. 

4 

1 B 
1 6 

7. 

8 

1 5 
16 



^tV 



4 


.683 


.655 


5 
8 


^tV 


^565 


546 


'f 


•571 


.673 


0. 
4 


'H 


.556 


559 


r7 


.567 


.632 


7. 
8 


2 


.562 


550 


2 


.562 


.601 


1 5 
1 6 


^iV 


'545 


536 


4 


'559 


.577 


I 


-i 


,529 


526 


4 


.556 


.558 


'A 


^i 


.528 


504 


4 


•553 


•543 


'i 


^f 


.526 


487 



Double Riveted Lap Joints. 



3 

8 


1 1 

1 6 


4 


.738 


754 


3 

4 


^1 • 


7 
16 


3 

4 


5f 


727 


734 


1 3 
16 


^f ■ 


1 
2 


1 3 
1 6 


4 


717 


721 


7 
8 


^H • 


9 
16 


i 


3 


708 


713 


1 5 
1 6 


^i • 


5 
8 


1 
16 


•> 8 


700 


707 


I 


3 


1 1 
1 6 


I 


3i 


692 


703 


'tV 


^ 8 


f 


'iV 


3A 


691 


688 


H 


^ 4 


1 8 
1 6 


'i 


^16 


684 


687 


^fV 


^ 8 


7 
8 


^A 


^ 3 


683 


675 


'i 


^ 2 


1 5 
16 


■i 


^ 8 


677 


676 


'1^ 


3t • 


I 


^A 


4 


672 


676 


'1 


3f • 



.714 

•705 
.689 

.674 

.667 
.660 

•654 
.648 

•643 

.638 
•633 



.718 

.690 

.684 

.683 

.670 

.660 
.652 
.646 

.641 

•637 
•634 



54 

It will be ob.served that values of Cc have been omitted. 
The crushing strength of iron and steel when confined and 
entirely filling a rivet hole has not been determined with 
accuracy, but it is so great that joints seldom fail through 
crushing. 

For butt joints single riveted, the following empirical 
equations give good results: 

d = t -|- K ^<^^i" how and iron rivets. 

d =r t + Y»^- for steel plates and steel rivets. 

For butt joints double riveted: 

d r= t + -A- l^or iron plates and iron rivets, 
d = t -]- ]- for steel plates and steel rivets. 

Prob. 52. Find the proper pitch of rivets for equal ten- 
sile and shearing strength in the double riveted butt joints 
of a soft steel boiler with steel rivets, the plates being one 
inch thick. Ans. 5.17 inches. 



Art. 28. vStkkngtii of Boii.krs. 

Equation [33] P = 7 niay be applied directly to 

the solution of problems connected with the strength of 
boilers whenever the efficiency e of the joint is known. 
Thus the total pressure tending to rupture a boiler length- 
wise is p 1 D, (Art. 22). One half of this pressure will be 
resisted by the riveted joint, the other half by the material, 
assumed to be at least as strong, on the opposite side. 



55 
Hence P =: — , and substituting in the above we have 

P D = J [34] 

The efficiency, however, depends upon the thickness of 
the plates and diameter and pitch of the rivets; or in other 
words, often depends upon the quantities we wish to find. 
In this case an indirect method for solving the problems 
must be used. For example, suppose we wish to find the 
proper thickness of a single riveted lap joint iron boiler 
whose diameter is to be 50 inches, and which mUvSt bear a 
pressure of 100 lbs. per sqr. inch with a factor of safety 
of 4. 

-TA .• / N . 4 X 100 X 50 0.212 ^ 

From equation (34) t = ~ ~ — =: . Turn- 

^ ^^^^ 2 X 47,000 X e e 

ing to Table XI, we see that the efficiency ranges from 0.583 

to 0.543. Assuming 0.56 as an average value, t = 0.4 nearlj^, 

or between ^ and JL inches. Taking from the Table 0.57 

lb ^ 

as a closer value for e, t = 0.372 or nearly ^ of an inch. 

Taking 0.571 as a still closer value for e, we have finally 

t = ^ of an inch as an answer to the problem. From the 

table, the corresponding diameter of rivets is ^ of an inch 

and the pitch 1% inches. 

Assuming this thickness, let it be required to find the 

factor of safety against transverse rupture, the transverse 

joints having rivets of equal diameter and pitch. The value 

7t D^ p 

of P in this case is and the length of a row of rivets 

4 
1 = ttD. Hence equation [33] becomes 

TT D^p TT DtSie ^ 4tS'te 
^ — or pD — 



f -- r- f 



56 

Inserting values and solving 

f = 4 X 50 X 47-000 X 3 X .571 ^ g 

100 X 50 X 8 '^^ 

or as before stated, (Art. 22) twice as great as for rupture 
lengthwise. 

The factor of safet}' is often made equal to 4 in practice, 
although sometimes taken as low as 3. 

Exercise. Describe processes of making solid drawn 
iron, brass, copper and lead pipes, and butt and lap welded 
pipes. 

Prob. 53. From the data given in Prob. 52, what is the 
efficiency of the joint, and what pressure would burst the 
boiler, supposing the diameter to be 10 feet ? 

Ans. 75.8 per cent; 902 lbs. per sqr. inch. 

Prob. 54. The diameter of a single riveted iron boiler 
with lap joints is 42 inches, the thickness of shell being ^ 
of an inch and the steam pressure 125 lbs. per sqr. inch. 
What is the probable factor of safet}^ ? Ans. 3.2. 

Prob. 55. A steel boiler is to have a diameter of 8 feet 
and to carry a pressure 200 lbs. per sqr. inch. The lap 
joints are to be double riveted and the factor of safet}^ is to 
be 4. What should be the thickness of plates and diameter 
and pitch of rivets? Ans. 14; i-^; 3I. inches. 

Prob. 56. An iron stand pipe or water tower is to be 30 
feet in diameter and 150 feet high. The joints are to be 
lapped with double rows of rivets, the factor of safet}' 
being 3. Find the proper thickness of the lowest and half 
way plates. Ans. 4-4- and ^ inches. 

^ -^ 1 b 1 u 

Prob. 57. In the last example what head of water would 
burst the stand pipe ? Ans. 463 feet. 



CHAPTER V. 



BEAMS AND CANTILEVERS. 
Art. 29. Reactions on the Supports. 






Fig. 22. 



Fig. 23. 



TT 



Fig. 24. 



P 



"7S 

R 



The distinction between a beam and a cantilever is that 
the former is supported at two or more places while the 
latter is supported at only one. 

Thus Fig. 22 represents 
a beam resting upon the 
supports Ri R2, and Fig. 

23 a cantilever, assumed 
to project horizontally 
from a wall, while Fig. 

24 shows a double canti- 
lever. 

If, as in Fig. 22, the 
supports are at the free 
ends, it is called a simple 
beam, but if either or both 
ends are fixed, it is a re- 
strained beam, and if rest- 
ing upon more than two 
supports it is a continuous 
beam. The space between 
supports is called the span. 



Py 



58 



The pressure of a beam upon a support is resisted b}^ an 
equal force acting in an opposite direction called the 
7-eactio7i. Usually the reaction acts upwards as in the above 

p figures, in which case the 



J 



P 



sum of the reactions must 
equal the sum of the loads, 
or the forces acting up- 
wards must exactl}' equal 
those acting downwards. 



S 

Fig. 24a. 

Sometimes, however, a reaction is downward as in Fig. 
24a, or it may be in any direction; but in any case, since 
the beam is assumed to be at rest, it is obvious that 

(y4) The sum of the foi'ces acting in any direction tnust 
exactly equal the sum of those acting in the opposite direction. 

Thus, as in Fig. 22, let the beam whose length is 1 inches 
be loaded uniformly with w pounds per inch of length, 
and also sustain a concentrated load of P pounds at some 
intermediate point. The total load acting downward is 
wl -j- P pounds. This must exactly equal the sum of the 
reactions Rx and R2 , or Ri -|- R2 = wl + P. 

Again, in Fig. 24a, in addition to the load wl + P, 
the reaction R2 also acts downward; hence 



R, = wl + P + R2 








Fig. 25. 



To make the illus- 
tration general, let 
the forces shown 
in Fig. 25 make 
angles with the 
vertical denoted by 

Oil OC2 ^3 /?! /?2 A 

etc. Resolve each 
force into its verti- 



59 

cal and horizontal components. Then since the sum of 
forces acting in opposite directions must equal each other, 
we have 

Pi cos OL^ -f P2 cos A'2 r= Rr COS A + R2 COS /ia -}- P3 COS OL^ 

Pi sin a'l -|- P3 sin a.^ + Ri sin fi^ = P2 sin a^ -\- R^ sin ^2 

The moment of a force is measured by the product of the 
force and its lever arm; or the force multiplied hj the 
perpendicular distance from the center of moments to the 
direction in which the force acts. 

Thus if we wish to measure the 
tendency of the force P, Fig. 26, to 
turn about the center A; or to find the 
moment of P with respect to the center 
of moments A, draw AB perpendicular 
to the direction of P and, calling this 
moment M, we have 

M = P X AB ==P a ^^^- 26. 

If a beam is at rest the tendenc}'- to turn about any point 
is zero or 

{B) The stun of the moments tending to tiir'^ii a beam in 07ie 
direction mtist exactly eqtcal the moments tending to turn it in 
the opposite direction. 

This is obviously true no matter where the point or 
center of moments is taken. A moment is called -|- w^hen it 
tends to produce motion in the direction in which the hands 
of a clock move, and — when in the opposite direction. 

The center of gravit}^ of a body is a point through which 
the resultant of the weight of all the particles always 
passes; or at which in finding the moment we may assume 
the entire weight concentrated, hence 




6o 

(O The DiGinent of a load is equal to the product of the load 
and the shortest distance from the center of moments to a vertical 
line passing through the center of gravity of the load. 

If the load is uniformly distributed, this line will pass 
through a point midway. 

Consider now a 
beam as uniformly 
loaded from end to 
end with w pounds 
per inch or foot of 
length, overhang- 
ing the supports by 



f- m-^ ( • - -^- n 



TT^ 



R 






1 ^V 
'3 



1 '^'1 

Fig. 27. 

distances m and n, and sustaining concentrated loads Pr 
P2 and P3 as shown in Fig. 27. From law (A) we have 
Ri -L R2 =: w m + w 1 + w n + Pi -f P2 4- P3 z= W. 
Since law (^) is true, no matter where the center of mo- 
ments is taken, let us choose the right hand support. The 
plus moments about this point must equal the minus mo- 
ments; hence with the aid of lav>^ (C) 

p. n = w mi r 1 -I- w 1 r Pi ai -|- P2 a2 



Ri 1 -f w n — 
2 



w m i r 1 -f w 1 

\2 / 2 



It will be observed that the reaction R2 does not appear 
in this equation because its lever arm is zero. In fact this 
point was chosen as the center of moments in order to avoid 
its introduction. All the quantities excepting R, , are 
supposed to be known; hence solving the equation, we 
have 

wm^ \yP wn^ 

1- w 1 m i f Pi ai -j- P2 a2 — P-, n 

Ri:rz_^ 2 2 

1 



Let us now take the center of moments at Ri 



In this 



6i 

case, for the sake of illustration, we will do what might 
have been done in the preceding, consider the entire uni- 
form load as concentrated at a point midwa}^ between the 
ends of the beam, thus 

R3 1 = w (m -f n -1- 1) X r^ + i^+^ _,^ _f_ p^ (J, _|_ 1) 

-|- P2 ( 1 — aa ) + Pi ( 1 — ai ) from which R2 can be found. 
If now we add Ri and R2 we should get the entire weight W 
as in the first equation and in this way prove the work 
correct; or, if sure no mistake has been made in finding Ri , 
we can evidently find R2 directly 

R2 z= W — Ri . 
For example, take a simple beam uniformly loaded, and 
also with a concentrated load P at a point one quarter of 
the distance between Ri and R2 

R, 1 = w 1 — + P 
2 

Ra 1 = w 1 -^ + P -^ or R2 == ^^-^ 4- ^^ and 

Ri -\- R2 = wl + P == W in agreement with law (A) 

Again in Fig. 24a let the entire length of the beam be 20 
feet, the distance betv/een Ri and Ra be 10 feet, the uniform 
load be 100 pounds per foot, and the concentrated load at 
the end be 1,000 pounds. Taking the center of moments at 
Ra we have 

Ri X 10 ^ 1,000 X 20 -h 100 X 20 X -^^ = 40,000 (lbs. ft.) 

or Ri =z 4,000 lbs. 
Taking center of moments at Ri 

Ra X 10 = 1 ,000 X 10 -|-- ico X 20 |- 10' = io,ooo(lbs.ft. ) 

or Ra = 1,000 lbs. 



^^ orRi 
4 


_ w 1 , 3 p 
2 ' 4 


— or R2 
4 


w 1 I p 

" 2 ' 4 



62 

The entire load on the beam is 3, coo pounds and since 

Ri is greater than this, Pvo must act downward. 

Three men carry a stick of timber of uniform section, 

, one taking hold at the end 

I . and the other two at some 

S ~7^ intermediate point. How 

2 R R far from the end should they 

T-A ^ be in order that each may 

bear an equal weight: 

Let X, Fig. 28, be the unknov/n distance, 1 the length 

and w the weight per foot. Taking the center of moments 

at the right hand end 

^ w P , , ^ w 1 , 3 1 

2 Rx = , but R = , hence x = -^ 



23 4 

Prob. 58. A beam carrying a uniform load of 850 lbs. 

per linear foot overhangs one support a distance of 6 feet, 

the span being 20 feet. Find the reactions. 

Ans. 14,365 and 7,735 lbs. 

Prob. 59. A simple beam with a span of 16 feet is 
uniforml}^ loaded with 900 lbs. per linear foot and carries a 
concentrated load of 2,500 lbs., 6 ft. from the left end and 
another of 2,500 lbs., 4 ft. from the right end. Find the 
reactions. Ans. Ri = 9,387.5 lbs; R2 = 10,012.5 ^^s. 

Prob. 60. A simple beam 20 ft. long weighs 100 lbs. per 
linear foot, and also carries a load of 2,500 lbs. Where 
should this load be placed in order that one reaction 
should be double the other ? Ans. 4 ft. from support. 

Prob. 61. In Fig. 25, let the forces Pi P2 P3 and Ri be 
3,000, 200, 1,000 and 500 lbs. respectivel}-; and the angles 
be each 30°. What should be the magnitude and direction 
of R2 to hold them in equilibrium ? 

Ans. R2 == 2,605 lbs.; A = 55° 3^'- 



63 

Prob. 62. If in Fig. 25 the distance of the point of 
application Pi (3,000 lbs.)) P2 (1,000 lbs.) and P3 (500 lbs.) 
is 10, 20 and 15 ft. from the point of application of Ri 
what is the turning moment of each force about this 
point, the angle being 30° in each case. 

Ans. -|- 25,980; -|- 3,464; — 12,990 lbs. feet. 

Prob. 63. Assuming the answers given in the two 
preceding problems, find distance A B of the point of 
application of R2 that the moments may be in equilibrium, 

Ans. AB = 1 1. 17 feet. 

Prob. 64. Three men carry a conical stick of timber 
whose diameter at the butt is d ft. and whose length is 1 ft. 
One lifts at the butt and the other two at some intermediate 
point. At what distance from the butt should they take 
hold that each may bear an equal weight ? (From base to 
center of gravity of a cone = ^ 1.) Ans. x :== ^ 1. 

Art. 30, Verticai, Shear. 

The difference between 

the sum of all the upward ^^ 1 

and the sum of all the j^ m -^ { } ^n ^ 

; ( """^ 1 

downward forces to the A 1 ^ " — ^ 



right or left of any as- ijl tl|p, ^^^ Jp *,q ^% 

sumed section of a beam 

is called the vertical shear. Fig. 29. 

Thus to the left of the section AB, Fig. 29, the only 
upward force is the reaction Rj The sum of the downward 
forces is wm -f wx -\- Pi + P^ and the difference, or 
vertical shear 

V = R, — w (m + x) — <^ P 

the last term standing for the algebraic sum of all concen- 
trated loads to the left. 



64 



Since the sum of all the upward forces must exactly 
equal the sum of all those downward, this quantity V rep- 
resents the shearing stress in the beam at the section con- 
sidered. Or in other words, if the right hand portion of 
the beam were removed, a vertical force of V pounds ap- 
plied at the section, would replace the stress and hold the 
beam in equilibrium. 

It is immaterial whether the forces to the left or right 
are considered, the value of V being the same in each 
case; but for the sake of uniformit}^ unless otherwise 
stated, only those to the left will hereafter be considered, 

Thus assuming a 



±1^ 



h-h 



A" 



I 



li.J 






iB 1 





r^ 


-i 


1 1 

1 


C 




E 


1 

1 






I 




G 




1 



simple beam, shown 
in Fig. 30, let us 
find the vertical 
shear at the section 
AB. First find the 
reactions, 



D 



H 



Ri 


= 


wl 

2 


1 


3P 
4 


R. 


:zz 


wl 


+ 


p 



Fig. 30. 
cal forces to the left of AB, 
3wl 



V = Ri 






— P = 

wl 



wl 



•h 



pounds. 



Taking the 
3 wl 



4 
verti.. 

P or 



V 2 4 '' 

Taking' the vertical forces to the right. 



w 1 



V 2 



+ 



P^ 

4 



The numerical value is the same in each case, the differ 



65 

ence in sign simply indicating the direction in wliich tlie 
force should act to replace the stress in the beam. If x is 
the distance of a section from the left support; for values 
of X greater than ^ 1. 

,. wl ,3? ^ -p wl P 

V = -+- — — — w X — F =z — — w X 

24 24 

and when x is less than }i 1. 

\v 1 3 P 

V =■ — — - -f- — wx pounds; and when x "== o, 

24 

T^ W 1 , ^ P ^ 

V = Ri =^ -f- -^ pounds. 

24 

The change in vertical shear ma}^ be graphically illus- 
trated as in Fig. 30. Let CD represent the length of the 
beam, and adopting any convenient scale, lay off CO as 
the value of V when x =: o. When x nearly = ^ L 

w 1 3 P w 1 w 1 3 P 



24444 

Make KF equal to this on the same scale, and join O 
and F. When x is a little greater than % 1, 

244 44 

which is plus or minus as — is less or greater than 

4 4 

P 
Assuming — to be greater, lay off KG- equal to — V. 

4 

\\rv, . , ^r -, wl P - , wl P 

When X — 1, \ =. — wl =^ 

24 24 

Making DH equal to this, and joining H and G, the fig- 
ure is complete, and the vertical shear at any section in 
the beam can be found by simply measuring the corre- 



66 




Fig. 



sponding ordinate below. Directl}^ under the load P, it 
will be observed that the shear suddenh' passes from plus 
to minus, or passes through zero. 

Fig. 31 represents a 
cantilever uniformly load- 
ed, and also with a load 
P at the free end. At any 
section distant x from the 
end, V = — P — wx 
When X = o, V = — P 
and when x = 1, 

V = — P — w 1 
Accordingly, adopting 
an)" convenient scale, la}' off EF = — P, and DH = 
— P — wl, and joining F and H, we have as before a 
diagram from which the vertical shear can be measured 
for any section. The reaction at the wall must be the sum. 
of the loads P and wl. Hence just inside the wall, 

V = — P — w 1 + P + w 1 =r o 
It is important to observe that in ever)^ case there is 
some section at w^hich the shear becomes equal to, or 
passes through zero, for it will be shown later that this is 
usually the dangerous section. 

Prob. 65. A simple beam with a span of 16 feet carries 

concentrated loads of 1,000, 2,000 and 3,000 pounds at 
distances of 5, 10 and 15 feet from the left end. Find the 
vertical shear just to the left of each of these loads. 

Ans. + 1,625; + 625; — 1,375, 

Prob. 66. Construct a diagram showing the variation of 
the shear in Prob. 65. At what section does the shear pass 
from + to — ? Ans. Under second load. 



67 

Prob. 67. A cantilever 20 ft. long is uniformly loaded witli 
600 Ibe. per linear foot and sustains also a ^veiglit of 1,000 
lbs. at the end, and another of 2,000 lbs. 10 feet from the 
end. Find the vertical shear at points 5, 10 and 15 feet 
from the end. Ans. 4,000; 9,000; 12,000 lbs. 

Prob. 68. Construct a diagram showing the variation 
of the shear. At what section does the shear pass from -|- 
to - ? 

Prob. 69. A beam carr3dng a uniform load of 1,000 
pounds per linear foot overhangs the left support a dis- 
tance of 10 feet, the span being 20 feet. Findt the sections 
where the shear passes from -\- to — . 

Ans. 7.5 feet and 20 feet from right support. 



Art. 31, Bknding Moments. 

The difference between the moments of the forces to the 
right or left of any section A B tending to turn a beam 
about this section (Fig. 29) is called the be7iding mo^nejit. 
This causes stresses in the material which must be re- 
sisted, and is the chief cause of rupture in beams. Thus 
taking the forces to the left of A B in the Fig,, and making 
right-handed m^oments + and left-handed — , the bending 
moment at A B is 

M = R, X — w(m-fx)-5L±JL — P, (m + x) — P3 (x — a) 

or, the bending moment at any section is equal to the moment 
of the 7'eaction minus the 7noments of the loads to the left. 



68 

If the forces to the right of the section had been taken, 
we should have the expression, 

M =: — R, (1 - x) -!- w(n + 1 - X) ^^ + ^ — ^ 

2 

But these two equations, as may readih^ be proved by 
finding and substituting the values of Rj and R2 are ex- 
actly the same, with exception of the sign. 

Thus, assuming the simple beam shown in Fig. 30, let 

us find the bending moment at the section AB. Taking 

moments of forces to the left, 

^r T. / 3I v^wl 3I PI /wl , 3P13I 

4 4 8 2 2 4/4 

9wP PI , . ^, 3wP , PI 
or, reducing M = — -\- 




32 2 ' 32 16 

Or, taking moments to right of the section, 

4 ^48- .2+4 

AX /3wP , PI 
or M = — \~ - 

\ 32 16 

which numerically is the same. The first result being + 
shows that the resulting moment tends to turn the left hand 
portion of the beam in the direction of the hands of a clock 
about the section, thus compressing the fibers on the top 
side and extending those on the lower. While the second 
result being — indicates a resultant moment of an opposite 
character for the right hand portion of the beam, also 
tending to compress the upper fibers and extend the lower; 
It is therefore immaterial whether the moments of the forces 
to the left or right are taken, a good rule being to solve 
each problem in the simplest manner. For sake of uni- 
formity however, the left hand moments will be taken 
unless otherwise noted. 



69 



Let X be the distance 
of any section AB, Fig. 
32, from the left sup- 
port. Then for all val- 
ues greater than a the 
bending moment 



H 



— ^A 



M 



Rt X — 




j^ 



F 

P(x — a) Fig. 

When X = a this becomes M = Rr a — >^ w a^ When 
X = 1, M = o. When x is less than a, M = Ri x — 
>^2wx^ and when x = o, M = o. 

The moments can be graphically shown similarly to 
vertical shear. Thus let C D represent the length of the 
beam, and assuming any convenient scale lay off ordinates 
to represent the moments corresponding to assigned values 
of X. In the above illustration M is zero both when x =: o 
and when x = 1. The curve then passes through the 

points C and D. When x r=: a, make E F = Ri a 

..r-u 1 1 r-TT ^^1 wP _/l 

When X = — , make GH = -— — — P — 

2 2 8 \2 

and in this way finding a sufficient number of points, draw 
the curve through them 
The bending moment at j^"" ^ "^ 
any other section can then 
be found by measuring 
the corresponding ordi- 
nate below. 

In the case of a canti- 
lever uniformly loaded, 
and with a load P at the Fig. 33. 




70 

free end (F'ig. 33) the bending moment at an}^ section 
distant x from the free end is, 

w x~ 
M = — — Px. The value of M will evidently 

be greatest at the wall, and will become zero at the free 
end. Intermediate values can readily be found b}^ substi- 
tution, or graphically as shown in the Fig. 

Prob. 70. A simple beam is uniformly loaded with 1,500 
lbs. per linear foot, the span being 15 feet. Find the bend- 
ing moments at the center and quarter points. 

Ans. 31,640; 42,188; 31,640 lbs. feet. 

Prob 72. In the last problem, assuming the beam to 
sustain weights of 8,000 and 4,000 lbs. at points 7 and 12 
feet from the left end, nnd the bending moment at the 
point of application of the last named load. 

Ans. Ri := 16,316,6, R2 = 18,183.3, ^I = 54o50 lt)S. feet. 

Prob. 73. A cantilever, whose length is 12 feet, is uni- 
formly loaded with 600 lbs. per linear inch. Find the bend- 
ing moment in pound inches at the fixed end. 

Ans. 6,220,800 lbs. inches. 

Prob. 74. In the last problem, what weight at the free 
end would give a bending moment of 160,000 lbs. inches at 
a distance of 6 feet ? Ans. 90 pounds. 

Prob. 75. The span of a simple beam being 10 feet, 
where should a weight be placed in order that the left 
reaction may be tvv'ice the right reaction; and what must 
be the weight in pounds to give a bending moment at the 
center of 500,000 lbs. inches, the weight of the beam being 
neglected? Ans. 40 inches from left end; P = 25,000 lbs. 



71 



Art. 32 Maximum Bending Moments. 



In the beam, illustrated by Fig. 32, there is obviously 
some section between the supports where the bending mo- 
ment is greatest, and in practice it is of much importance to 
determine this accurately, since here it is that the beam is 
most likely to break. 

From the curve it will be seen that where the tangent TT 
to the curve is horizoiital the ordinate representing M is gixat- 
est. If then we can find the value of x for this tangent 
point, we shall know where the dangerous section is, and 
b}^ substitution find the maximum bending moment. Let 
T, Fig. 34, be any point 
on a curve showing the 
relation between M and x, 
and through T draw any 
secant line SS cutting the 
curve also at a, and let (^ 
be the angle between SS 
and the axis of X. From. 

a b 



the Fig. tang, p =. 
Mx — M 



X, 



X 



bT"" 
but if the beam 




Fig. 34. 



is loaded as in Fig. 32, M = Ri x -- 



wx^ 



- P ( X — a ) 



and Mi = R^ Xi 



W Xi 



P ( Xx - a ) 



hence tan. /? = 



Rx (Xx— x) 



w 

Xt - 



X2 ) — P (Xx— X) 



X 



72 

or dividing b}^ Xi - x, tan ,6 = R^ (Xi + x) — P 

If new we revolve the secant SS about the point T, the 
point a will approach T, and when Xi becomes equal to x. 
the secant will become tangent to the curve, for it will then 
touch it in onl}^ one point. Hence making Xj = x in the 
above equation and calling the angle alpha, we have 
tan cy = Ri — w X — P, 

a being the angle the tangent line makes with the axis of X 
at any point in the curve whose absissa is x. 

At the point where the tangent itself is horizontal or 
where the bending moment is greatest, the angle a becomes 
equal to zero. Hence for this point we have the important 
relation, 

o =z Ri — w X — P or X = 

w 

Let us now find an expression for the vertical shear at a 
distance x from the left support. From Fig. 32 

V = Rx— wx — P. 
But this is exactly the expression derived above. Hence 
we can conclude that 

T/ie niaxhmnn bending moment occurs at a pohit ivhere the 
vertical shear equals or passes through zero. 

Thus in Fig. 32, let the span equal 20 feet, with a 
uniform load of 500 lbs. per foot, and let the load P of 1,000 
lbs. be applied at a distance of 5 feet from the left end. 
Taking the center of moments at the right support 

20 Ri = 500 X 20 X 10 4- 1,000 X 15 = 115.000 lbs. ft. 
or Ri — 5,750 lbs. 

The vertical shear just to the right of the load P, is 
V = Ri — 5 w — P = 5,750 — 2,500 — 1,000 -= 2,250 lbs. 



73 

Since this is greater than zero the maximum bending 
moment must be to the right of the load P. Taking x for 
the required distance, \yq have 

V = Ri — wx — P = 4,750 — 500 X 

and since this must equal zero, 500 x = 4750 or x = 9.5 ft. 

Substituting this value of x in the equation 

M — R,K — ^^^ — P (x - a) 
2 

we have finally, Max. M. 1= 27,562.5 lbs. ft. 

It will be observed that we first decided whether the 
shear would become zero tinder the load; for if this had oc- 
curred the formula V = Ri — w x — P =: o could not 
have been used, since it only holds true for points to the 
right of the load. 

Thus suppose the load P to be 16,000 lbs. In this case 
Ri = 17,000 lbs. and adopting the above formula 

V = 17,000 — 500 X — i6,ooc = o 

we find X = 2 ft. But the equation is only true for values 
of X greater than 5 ft. Hence the shear must equal zero 
either under or to the left of P. To the left V :=i 17,000 
— 500 X and making this equal to zero, x = 17,000 -^ 5000 
= 34 ft. or in other words, the load P needs to be 34 feet 
from the left end in order that the shear becomes equal to 
zero before reaching it. Just to the left of P the shear is 
17,000 — 500 X 5 = +14,500 lbs. Just to the right it is 
17,000 — 500 X 5 — 16,000 = — 1,500 lbs. Hence it 

becomes equal to zero or passes /rom minus to plus under 
the load, showing the dangerous section to be at this point. 



74 

Prob. 76. Prove that when W is the entire load, the 
maximum bending moments are : 

A cantilever loaded at the end, M ;= W 1 



A cantilever uniformly loaded, M =^ 

A simple beam loaded at the center, M 1= 
A simple beam uniforml}^ loaded, M = 



W 1 

2 
W 1 

4 

Wl 

8 



Prob. 77. A simple beam whose span is 20 feet, carries 
loads of 2 tons, 4 feet from each support. What is the maxi- 
mum bending moment, neglecting the weight of the beam ? 

Ans. 16,000 lbs. feet. 

Prob. 78. In the last problem, assume also a uniform 
load of 1,000 lbs. per linear foot, and find the greatest 
bending moment. Ans. 66,000 lbs. feet. 



Art. 33. General Expressions for the Bending 
Moment and Vertical Shear. 

Let a beam, Fig. 35, 
I I , 1 extend beyond the left 



i 



/^p ' p A support a distance m, and 

^^ "^ ^ ^ besides a uniform load, 

-pj^ ^„ carry concentrated loads, 

represented by P, whose 
center of gravity is at a distance a from the support. 
Then, since the bending moment is always equal to the 



75 

difference between tlie moments tending to bend the beam 
in opposite directions, at any distance x 

M = Rx X — w (m + X) ^ "^ ^' 

wni^ .. _ 
=: Ri X — w m X — Pa — P x 

2 2 

wm^ , ^ \ , ,^ ^. wx' 
\~ Pa -r (Ri — w m — P) x 




/ ^y m^ \ 

But from the Fig. we see that — ( — f- P a 1 is the 

bending moment at the left support, and (P.i — vv^ m — P) 
is the vertical shear just to the right of the support. 
Calling this moment M', and the shear V' w^e have 
generally 

M = M' + V X - ^ [35] 

or in words: 

The bending moment at any point is equal to the bending 
moment at the left support^ plus the moynent of the vertical shear 
just to the right of this support, rniiius the moments of the loads 
between the suppoi't aiid the point. 

This general law" finds important application in beams 
fixed at one end or resting on more than two supports. 

From Fig. 35 it will be seen that the vertical shear at the 
section distant x from the support is 

V = Ri — P — w m — w X or 

V = V'' — v/x [36] 

or in words: 

The vertical shear at any point is eqiial to the vci'tical shear 
just to the right of the left siipport^ mi?ius the sum of the loads 
between this support and the point. 



76 




Thus let Fig. 36 represent 
a beam uniformly loaded 
and overhanging each sup- 
port equally. At any dis- 
tance X between supports 

w m^ ^ 
and V = 



wl 



\v m 



Hence 



Between the supports 



w X and making this equal to 



zero, we see that the dangerous section is where x = 
Substituting this in the expression for M, we have 



Max. M =z 



w m^ w P 



w V' 



w m 



4 



- + 



2 * 
wP 



Over each support the shear again passes through zero, in- 
dicating that at these points there are maximum moments. 



w m^ 



Thus at the left support it is M 



By inspection of the Fig. it will be seen that the tendency 
of the load is to bend the free end of the beam downwards, 
making the upper side convex, while at the center the 
reverse is true, the curvature being concave. Hence 
between supports there must be points where the beam is 
not bent at all; or in other words, where the bending mo- 
ment must be zero. These are called inflection points^ and 



77 

tliey can readily be found b}' placing the general expres- 
sion for M equal to zero, and finding the corresponding 
values of x. 

Thus in the above, making 

\v m^ w 1 X vv x^ 

M = = o 

2 2 2 

we have the quadratic x^ — 1 x =^ — ni- . Solving, 

x == — ± \ — — ni^ = — ±: — -X 1 — 4 n 

2 \ 4 2 2 \ 



-.2 



Prob. 79. A beam uniformly loaded with i,oco lbs. per 
foot, overhangs the left support a distance of 10 feet, the 
span being 20 feet, and carries also a load of 2,000 lbs. at 
the end, and another of 2,000 lbs. midway betv/een the sup- 
ports. 

(a) Find the reaction. Ans. Ri = 26,500; R2 = 7>500 
(d) The maximum moment between the supports. 

28,125 l^s- ^^^t 
(r) The maximum negative moment. 70,000 lbs. feet 

Prob. 80. A beam, uniformly loaded, overhangs each 
support an equal distance m, the span being 1. Find the 
relation between m and 1 when the maximum positive and 
maximum negative moments numericall}^ equal each other. 

Ans. m = 1 -^ |/8. 

Prob. 81. Assuming the answers given to Prob. 79, find 
the inflection points. Ans. 6.12 feet from left support. 



CHAPTER VI. 
THE STRENGTH OF BEAMS. 

Art. 34. The: Neutrai. Surface and Neutral Axis. 

We have seen in the previous chapter that loads upon a 
beam produce vertical shear and bending moments which 
must be resisted b}- the strength of the material. In the 
former case the formula V = A Ss can be directly applied, 
where A represents the cross section of the beam in square 
inches and Ss the unit shearing stress caused b}' the shear. 
The latter case is more complicated. Tnus let Fig. 37 rep- 
resent a cantilever bent b}'- a load P at the end. It will be 
seen at once by inspection that the fibers on the upper side 
are extended while those on the lower side are compressed; 
the length a b, originally equal, being now greater than c d. 
Between the upper and lower sides there must be a point 
where the fibers are neither extended or compressed, or 

where their length is unchang- 
ed, and consequently where 
there is no additional stress 
caused by the bending action, 
e and f are such points, and if 
through them and other points 
similarly found we pass a line 
O O it Vvdll lie in what is known 
as the NeuU'al Siu-face or a 
^IG. 37. surface in which the fibers are 




79 

not changed in length. A line at right angles to the plane 
of the paper and passing through an}^ point of 00, say at 
e, is the neutral axis for that section. 

We will now show that the neutral axis must pass 
through the center of gravity of each cross section. 

If through e we draw a line parallel to bd, and consider 
a fiber whose original length was ef , the upper shaded por- 
tion shows the increase in the length due to the bending, 
while the lower portion shows the corresponding decrease. 
The outermost fibers are stretched or compressed mcst. 
At the neutral axis the length is unchanged, and b}^ in- 
spection it will be seen that both above and below, the dis- 
tortion or stretch is directly proportional to the distance 
from the axis. From this it follovv^s, since by law (A) 
Chapter II, stresses are proportional to the elongations 
that the unit sty^esses due to bending are directly proportional 
to the distances from the 7ieutral axis. 

Let Fig. 38 represent a cross section 
of a bent beam and ee be the neutral 
axis. Designate the distances from 
this axis to the outermost fibers by c 
and Ci and the corresponding unit 
stresses by S and Si Let us now con- 
sider a fiber whose distance from the 
axis is x and where the stress is vSx . 

By the above law, -^ — = — or 

o c 

S X 

Sx = lbs. per sqr. inch; and if the area of the fiber is 

c 

a square inches, the total stress acting on it is a Sx = — 

pounds. Upon any other fiber whose distance is Xi and 




So 

S 3 X 

area aj the stress is -^-^^ — - pounds, and the total stress 

c 

at right angles to the section above the neutral axis is 
3 

—(a X + ai Xi + a2 X2 + etc.) pounds. Similarly, the to- 
tal stress upon fibers below the axis, v^^hose distances are 
y Yi 3^2 , etc., and areas b bi b2 , etc., is -— ^(b y + bi yj_ + 

bz y2 + etc.) pounds. But since, as shown in the last 
chapter, the forces acting in one direction must exactly 
equal those acting in the opposite, it follows that 

— (a X + ai Xi + a^ X2 -f etc.) = -^ (b y + bi 3-x + b2 3^2 ) 
c Ci 

But, b3^ the above law, r^- == — or — = — ~ Hence 

Oi Ci C Ci 

a X + ai Xi + a2 X2 + etc. = b y + bi yi + ba y2 + etc. 

Each of these quantities, ax, b 3^ etc. represents the 
statical moment of a portion of the area with respect to the 
axis ee; and since the sum of the moments tending to turn 
in one direction is equal to the sum of those tending to 
turn in the other, it follows that 

T/ie neut7'al axis rnust pass through the center of gravity of 
the cross section considered, and lie in the neutral surface. 

Methods of finding the center of gravit3^ properl3^ belong 
to elementar3^ mechanics, and its position for some of the 
more common sections will be given without proof at the 
end of Art. 36. Generally a section of an3^ given shape 
can be divided into rectangles and triangles in each of 
which the position of the center of gravity is known. The 
center of gravity of the whole can then be readil3^ found 
by use of the principle that the total area, multiplied by the 



8i 



distance from its center cf gravity to any axis lying in its 
plane, is equal to the sum of the moments of the component 
parts with respect to this axis. 

Thus, let Fig. 39 repre- 
sent a rectangle with a por- 
tion removed, and z the 
distance from the lower side 
to the center of gravity of 
the Fig. The total area is 
evidently b (h — hi) + b ho 
Adopting XX as the axis 
of moments, by the principle 
just stated, Vv-e have, 

[b (h — hi) + b h^] z — b (h — hi) 

from which z can readily be found. 

I^et Fig. 40 represent a 
trapezoidal section, its area 
b + bi 




being 



X h. The 



D b 


B 




We 




\\ 



center of gravity of a trian- 
gle is in a line parallel to 
the base, and one third of 
the altitude from it. Hence 
dividing the trapezoid into Fig. 40. 

two triangles by the line AB, we have the relation, 



A 



Bi 



(b + bi)hz 



bh 2h 
_ X — 



, bih h 

^ 3 



h^ / 



b + 



2 



h/2b -f bi 
3 V b + bi .^ 

If the figure is of such a shape that calculation becomes 
difficult, for practical purposes it is generally sufficient to 



82 



cut it out of stiff cardboard and experimental!}' find the 
center of gravity b}' balancing on a sharp edge. 

Prob. 82. From an equilateral triangle, the length of 
each side being 12 inches, a circle having a diameter of 4 
inches is cut, the center of the circle being on a line bisect- 
ing an angle and 3 inches from the opposite side. Find 
the distance to the center of gravit}' of the figure from this 
side. ■ Ans. 2.76 inches. 

Prob. 83. A four inch hole is bored at right angles 
through a 2 X 10 beam, the center of the hole being 2 
inches from the center of the beam. Find the greatest dis- 
tance from a side to the center of gravity of the beam. 

Ans. 6y3 inches. 

Prob. 84. A trapezoid, whose lower and upper sides are 
12 and 8 inches long, and whose altitude is 6 inches, is 
capped by a triangle whose altitude is 3 inches. Find the 
distance from the base to the center of gravity of the fig- 
ure. Ans. 3}^ inches. 



Art. 35. The Resisting Moment. 

As shown in the preceding article, a bending moment 

/ produces stresses in a beam 

/ which are proportional to the 

*^^ ' / distance of the fibers from the 

— — — O / 

/ neutral axis. Thus if c is the 

/ distance from this axis to the 

/ outermost fiber where the unit 

/ stress is S pounds per square 

inch, at anj' distance x the 

S x 

stress is Sx = lbs. per 

Fig. 37. c 




83 

square inch, and if the area or cross section of the fiber is 
a square inches, the total stress on it is ^— pounds. 

The moment of the stress with respect to the neutral axis 

Sax S a x^ 
is -^^^ X X = pounds inches. This is the i^esist- 

hig moment of the fiber. For other fibers it is 

^- — ^— ^ — ^, etc., and the total resistin,^ moment is 

c c 

the sum of all these, or 

Resisting moment = -^(a x- -j- a^ Xi^ + ^2X3^ + etc.) 

The sum of such a series as that within the brackets is 
known as the moment of inertia. 

In this case it is the 77i07nent of inertia of the cross section 
under consideration with j'-espect to the rieutral axis. It will 
hereafter be designated by the letter I, hence 

T. • .• S I 

Resisting moment = -p— 

But since the beam is in equilibrium, the resisting moment 
m,ust exactly equal the b ending 7noment which catises it. Hence 
the important relation 

M = ^ [37] 

This is the fundamental equation for the strength of 
beams and cantilevers. 

Since the dimensions of the cross section are usually 
taken in inches, the resisting moment will be in pounds 
inches. Hence the bending moment should also be in 
pounds inches. 



54 



x\kt. 



MorviENTs OF Inertia. 




Methods of finding moments of inertia are more properly 
explained in Mechanics, and involve considerable knowl- 
edge of higher mathematics. Results for the more common 
sections are given below without proof, and b}- aid of these, 
and by dividing into rectangles and triangles, the momj.ents 
of inertia of other figures can usually be determined. The 
following principle will be of great service. 

Let Fig. 41 represent any section of 
a beam, ee be an axis passing through 
the center of gravit}^ and YY any par- 
allel axis lying in the plane of the fig- 
ure. Denote the area of narrov^' strips 
Y^parallel to YY by a ai a, etc., their 
corresponding distances from YY by 
X Xi X2 etc., and let z be the distance 
Fig. 41. between the axes. Then if I is the 

moment of inertia of the figure about the neutral axis ee, 

I = a(x — zy + ai(Xi — Zi)2 + a,{x. — z,y + etc. 
or, squaring and arranging 

I = (ax^ + aiXi^ -i- aoXa" + etc.) — 2z(ax -f a^x^ + 
a., Xo + etc.) + z2 (a + ai + ao -f- etc.) 

But (a x^ + ai x^^ -h a,3 x./ + etc.) is the moment of iner- 
tia of the figure about the axis YY, which designate by V; 
and (ax + aiXi -f aoX. + etc.) is the statical moment 
of the figure about the same axis. This must equal the 
total area A multiplied by the distance of its center of 
gravity from the axis, or is equal to Az. Noting also 
that (a + ai -f a, + etc.) = A, the equation becomes, 

I =. r — 2Az^ + Az^ = r — Az^ [38] 



85 

Or in words, the mo?neiit of inertia about the neutral axis is 
equal to the vtoment of iuertia about any parallel axis, minus 
the product of the area of the section and the square of the dis- 
tance betwee7i the axes. 

Thus, again considering Fig. 39, and assuming the dis- 
tance z of the center of gravity as known, let it be required 
to find the moment of inertia of the section about the neu- 
tral axis ee. Considering first the upper portion, its mo- 
ment with respect to XX, (see Fig. 43), is — (h^ — h^^ ) 
The moment of the lower portion about the same axis is 

The total moment of inertia about XX is therefore 



bh,3 



I' = — (h3 — lv-j-h.^3). Xhe area of the figure being 
A = b(h — hi -\- h.), form.ula 58 gives, 

I - -^(h3 _ hi3 -f h,3 ) — b (h - hi H- h, ) z^ 
3 

from which I can readily be determined. 

Consider again Fig. 40, in which the area A and the 
distance z from the base to the neutral axis has been de- 
termined. The moment of inertia of the upper triangle 
ABD about AC is given below (see Fig. 45) as bh3 ~- 4. 
That of the lower triangle ABC is bi h3 -f- 12. The total 

b h3 b, h3 

moment of inertia is therefore, I = • -}- — = 

4 12 

h3 
(3 b + bi ), and the required moment about the neutral 

. h3 

axis e e IS I = ( 3 b -[- b^ ) — A z^ 

12 



S6 



lu the accompanying figures, the moment of inertia about 
a specified axis is given, and also expressions for the area 
and distance c from the outermost fiber to the neutral axis, 
this axis being designated by the line ee. 





i 


b 


-H 


e 

— ^ — 

c 

^x 


Axis ee. 

A = bh. 

_ h 

2 

^ b h^ 

T2 


Axis XX 


1 

-1- 

h 

1 


■ 


J. _ bh3 



Fig. 42. 




Axis XX. 

r = -^(h3 - 
3 



hi3 ; 



Fig. 43. 




Axis e e. 

A = bh — b;hi 
h_ 
2 



c = 



I = — (bh3 — bih,3 ) 
12 



Fig. 44. 



Axis ee. 
bh 

2 

2h 



A = 



3 

bh3 



Axis 


ee. 


A =: 


4 


c = 


>4d 


I = 


TTd^ 



64 



87 



Axis XX. 
bh3 



r = 



4 



r 



Axis YY. 
bh3 



12 




Axis XX. 



64 



Axis e e. 

A = — (d^ - 
4 

c = Md 



I 



64 



(d4 



d,0 



Fig. 45. 




Fig. 46. 




Fig. 47. 



13 



88 



^-b-•^l 




Ellipse. Axis ee. 

A = ^bh 
4 

c = >^li 

I = ;^bh3 
64 



^ 



Y 



Yl 




\ 



s 



^ 



Many other sections might be 
given, which however would be of 
but little real use. Commercial 
forms, such as "I" or "deck" 
beams, vary much in shape ; and 
manufacturers usually give mo- 
ments of inertia, areas, and weights 
in their trade catalogues. 

Fig. 49 represents a section of an 
iron " I beam," and Fig. 50, of a 
deck beam, the two forms most 
commonly used as beams. 
The following table is taken from the catalogue of Car- 
negie Bros. & Co. Other manufacturers produce similar 
beams, but with dimensions more or less changed. The 
last column gives values (IJ for the moment of inertia 
about the axis YY. 



Fig. 49. Fig. 50. 



89 







TA 


.BLE : 


XII. 














I Beams. 








Designation 
Depth 

in 
Inches 


Weight 

per 

Foot. 

Pounds 


Area 

of 

Section. 

Sqr. In. 


Thickness 
of Web 

Inches 


width 
of 

Flange. 
Inches 


I 


I 

c 


L 


15'' Light. 
15'' Heavy 
15'' Extra h 


67 
80 

50 


20.1 
24.0 
16.0 


0.67 

0.93 
0.47 


5-55 
5.81 

5-33 


677 
750 
614 


90.3 
1 00.0 

81-9 


25-4 
29.9 

20.0 


12'' Light 
12'' Heavy 


42 
60 


12.6 
18.0 


0.51 
0.96 


4.64 
5-09 


275 ■ 
340 


45-9 
56.7 


II. 
15-5 


10'' Light 
10'' Heavy 


30 

45 


9.0 

13-5 


0.32 
0.77 


4.32 
4-77 


150 

187 


30.0 
37-5 


7-94 
II-3 


g' Light 
9'^ Heavy 


23-5 
33-0 


7.0 

9.9 


0.26 
0.58 


4.01 
4-33 


97-5 
117 


21.7 
26.0 


5-48 
7.14 


9'' Extra L 
g'' Extia H 


45-0 
50.0 


13-5 
15.0 


0.75 
0.91 


4.94 
5.10 


159 
169 


35-3 
37-5 


14.0 

15-7 


8'' Light 
8'' Heavy 


22.0 
35-0 


6.6 
10.5 


0.31 
0.79 


.3.81 
4.29 


69.9 
90.4 


17-5 
22.6 


4-57 
6.96 


7" I^ight 
7''' Heavy 


18.0 
25.0 


5-4 

7-5 


0.23 
0.53 


3.61 
3-91 


45-8 
54-3 


15-1 
15-5 


3-72 
4.87 


6'^ Light 
6'' Heavy 


13-5 
18.0 


4.1 

5-4 


0.24 
0.46 


3-24 
3-46 


24-5 
28.4 


8.16 
9.48 


2.00 
2.51 


5'' Ivight 
5'' Heavy 


10. 
13.0 


3-0 
3-9 


0.225 
0.405 


2.73 
2.91 


12.3 
14.2 


4.94 
5-69 


1.08 
1-34 


4" Ivight 
4'^ Heavy 


8.0 
10. 


2.4 
3-0 


0.23 
0.38 


2.48 
2.63 


6.19 
6.99 


3.10 
3-50 


0.71 
0.87 


3'' Light 
3'' Heavy 


7.0 
9.0 


2.1 
2.7 


0.19 
0.39 


2.32 
2.52 


3-09 
3-54 


2.06 
2.36 


0.55 
0.84 



90 



Desiguatiou 
Depth 

in 
Inches 



Deck Beams. 

Weight Area Thickness Width 

per of of Web of 

Foot. Section. Flange. I 

Pounds Sqr. In. Inches Inches 



// 



Light 
Heav3^ 

Light 
Heav}" 

Light 
Heavy 



23-5 
30.0 

21.5 
28.0 

17.0 
23.0 



7-1 
9.0 

6.5 

8.4 

5-1 
6.9 



0.406 
0.625 

0.500 
0.750 

0-375 
0.625 



3-75 
3-97 

3-75 
4.00 

3-50 
3-75 



78.6 
91.9 

52.1 
63-3 

34-4 
43-0 



c 

17. 1 
20.0 

II. 6 
14. 1 

8.6 
10. 8 



Ii 

2.49 

3-17 

2.23 
2.96 

1. 81 
2.41 



When, however, an irregular form is used, the moment of 
inertia of which cannot be readily found from tables, or by 
direct calculation, the following graphical method may be 
employed. 

Ar- Let Fig. 51 represent a 

-.J^rail, the dimensions of which 
are knov/n. Make the line 
AB equal to the entire depth 
and draw any line, whose 
length is b inches, at right 
angles to it, the distance 
from the base CD being x 
inches. The moment of this 
Fig. 51. line w4th respect to CD is 

bx^. Assuming any convenient scale, lay off a c equal 
to bx^ . At any other point distant Xi inches from CD, 
where the width is bi inches, the moment is Pi xr . Lay 
off a^Ci equal to this, to the same scale, and determine the 
point Ci In a similar manner find a sufficient number of 
points to construct accurately the curve AcdCoB. By 
means of a planimeter, or in any other way, find the aver- 




D B 



91 

age length in inches of all the lines a c, aiCi aoC, etc. This 
length multiplied by the depth AB in inches, and b}' the 
scale adopted, will give approximately the moment of in- 
ertia I' of the whole figure about the axis CD. Or, what 
is the same thing, find the area of the figure Ac CiC^B in 
square inches, and multiply it by the scale. Then, know- 
ing the distance z from CD to the center of gravity of the 
section, and its area A, the required moment of inertia 
can be found by the equation I 1= I^ — A z^ . . 

Exercise i. Explain how the distance of the center of 
gravit}^ from the line CD can be found in a similar way. 

Exercise 2. Apply the above described graphical meth- 
od to the determination of the moment of inertia of some 
figure in which it can readily be calculated, and compare 
results. 

Prob. 85. Assuming the answer given to Prob. 82, find 
the moment of inertia of the figure about the neutral axis, 
and the value of c. Ans. I =■ 191. 5 c = 5.72 inches. 

Prob. 86. Assuming the answer given to Prob. 83, find 
the moment of inertia of the figure about the neutral axis, 
and the value of c. Ans. I = 102.3 c = 6^ inches. 

Prob. 87. Assuming the answer given to Prob. 84, find 
the moment of inertia of the figure about the neutral axis, 
and the value of c. Ans. I = 320. c = 5.5 inches. 

Art. 37. Modulus of Rupture. 

It has been shown that the effect of a load upon a beam 
is to cause compressive stresses on one side, and tensile on 
the other. From Table IV we see that ultimate strength 



92 

for compression and tension are not usually the same, the 
tensile strength of cast iron, for example, being 20,000 lbs. 
per sqr. inch, ^Yhile the compressive is five times as great. 
Experiments show that when a beam is loaded up to the 
breaking point, the greatest stress caused lies between the 
values given in the table. This ultimate stress is called 
ih^ Modulus of Rupture. It depends much upon the shape 
of the cross section, and can onh' be approximately esti- 
mated. It is usual therefore, in the absence of definite 
experiments, to take the least value given in Table IV as 
the ultimate strength, and use this in the formula M = 

vSI 

— , thus being always on the safe side. 

The follovv"iug table gives some average values deter- 
mined by experiments upon rectangular cross sections.* 



TABLE XIII. 
Ultimatk Strength in Pounds per Square Inch. 

Tension. Compression. Modulus of Rupture. 

Cast Iron 20,000 90.000 35,000 

Steel 100,000 150,000 120,000 

Wrought Iron 55>ooo 55. o<^ 55>coo 

Timber 10,000 8,000 9,000 

Stone 6,oco 2,000 

It will be seen that for the more im.portant materials of 
which beams are made, iron, steel, and wood, the modulus 
of rupture lies about midwa}- between the ultimate tensile 
and compressive strengths. 

* Merriman's "Mechanics of Materials," page 62. 



93 

Prob. 88. A simple wooden beam, 4 inches wide and 12 
inches deep, whose span is 16 feet, sustains a load P at the 
center. 

(a) Find the greatest unit stress when P = 2,000 lbs. 

Ans. 1,000 lbs. per sqr. inch. 

(b) What is the greatest load the beam could sustain 
with a factor of safet}^ of 6 ? Ans. 3,000 lbs. 

(c) What load would break it ? Ans. 18,000 lbs. 

(d) If a two inch hole were bored through the center at 
right angles to the beam, what would be the greatest unit 
stress when P = 2,000 pounds? 

Ans. 1,004.6 lbs. per sqr. inch. 

(e) What load would break it? Ans. 17,917 lbs. 



Art. 38. The Stresses in Beams. 

Let it be required to find the greatest stress caused by a 
load of W pounds upon the 



- t 1 



end of a cantilever w^hose 1 y. 

length is 1 inches (Fig. 52), iW7\ 

the cross section being rect- Y/^ 

angular. Our fundamental 

equation is «Aw 

M = or S = -V- Fig. 52. 

c I 

The dangerous section is at the wall, where M = Wl. 

The moment of inertia I =: , and c = — . Hence 

12 2 



94 



for a cantilever Vvitli a load at the end of W pounds 

q 6W1 ^, 

o = . ■■ . — lbs. per sqr. inch, 
b h' ^ 

Suppose the cantilever to be loaded with w pounds per 

inch of length. The maximum bending moment in this 

^vl' 

. rience for a cantilever uniformly loaded 



is 



AT 



S = 



w r- 



3 w 1 

-^ lbs. per sqr. inch. 



b h- b h- 

ALgain, suppose a beam of the same length and section 

to be supported at the ends, 
with a load of W pounds at 
"A the center, as shown in Fig. 
53. The dangerous section is 
at the center, and the greatest 

Wl 



k — 



1 



Fig. 53. 



bending- moment is M 



4 



Hence for a simple beam with a load W at the center 



3WI 



w, — , , 5- lbs. per sqr. inch. 
2 b h" 

Suppose the same load of W pounds to be uniformly dis- 
tributed over the beam, the load per inch being w pounds. 

ATA1 1 -, • • -» r R 1 ^v 1" vvP 

Ine ereatest benains: moment is M =. — =z — — 

^ ^ 248 

Hence for a simple beam uniforml}^ loaded 






-. W 1 



4bh 



J- lbs. per sqr. inch, 



We see, therefore, that the same load in each case would 
cause stresses varying as the numbers 6, 3, i}2y and ^; 
or the stress caused b}' a weight of W pounds at the end of 
a cantilever, would be eight times as great as that caused 
by the sam.e weight uniformly distributed over a simple 
beam of equal length. 



95 

To find the breaking load in each case, let us suppose 
the beam to be 20 feet long, 2 inches thick, and 12 inches 
deep, the modulus of rupture for the material being 9,000 
lbs. per sar. inch. 



[I] 


w = 


b h' s 
61 ■ 


_ 2X 144X9,000 _ 
6X20X12 ~ 


: r,8oo pounds 


[2] 


\V =3 


bh^S 
3I " 


=. 3,600 pounds. 




[3] 


w = 


2 b h^- S 
3I 


r= 7,200 pounds. 




[4] 


w = 


4bh^S 
-, 1 


=: 14,400 pounds. 





To find the depth h in order that, in each case, the beam 
may sustain a load of i,oco pounds with a factor of safety 
of 4: 

TtI h - / 6W1 . . 

I_ij n — ^w — =: 17.Q inches. 



[2] h r= -x -^ =^ 12.6s inches. 

V bS 

[3] h = -x h^^ = 8.64 inches. 

V 2bS 

[4] h = -^/ 3 = 6.32 inches. 

V 4bS 

Prob. 89. A cast iron cantilever 2 inches wide and 6 inches 
deep projects from a wall. 

(a) When its length is 10 feet and a load of 1,000 lbs. 
is hung upon the end, what is the greatest unit stress 
caused by this load? Ans. 10,000 lbs. 

14 



96 

(b) What should be the length that it ma)- break under 
this load ? Ans. 35 ft. 

Prob. 90. A simple beam is constructed of a heav}' lo^' 
I beam. 

(a) If the span is 20 ft., what total load uniformly dis- 
tributed would it sustain, the factor of safety- being 10? 

Ans. 343.7 lbs. per foot. 

(b) What uniform load would break it? 

(c) What should be the length between supports that it 
ma^^ break under its own weight? Ans. 174.8 ft. 

(d) A 5 inch hole being made at the center of the beam, 
what uniform load would it sustain, the length being 20 ft. 
and factor of safety 10? Ans. 328.1 lbs. per linear ft. 

Prob. 91. A floor is to sustain a uniform load of 200 lbs. 
per sqr. ft. The span of the floor beam.s to be 20 ft., and 
their distance from center to center is to be 3 ft. The factor 
being 6, what I beam should be selected? 

Ans. — = 39.3 Light 12''. 

Prob. 92. A simple beam w^eighing 40 lbs. per cubic ft., 
whose span is 20 ft. has a section at the center described in 
Prob. 83. Assuming the answers given in Probs. 83 and 86, 

(a) What uniform load W would break it? 

(b) What load P at the center would break it, leaving 
out the weight of the beam? Ans. 2,423 lbs. 

(c) What load P at the center would break it, taking 
into account the weight of the beam? Ans. 2,414.2 lbs. 



97 






-^ 



Prob. 93. A simple 2'''Xi2'''' wooden beam is construct- 
ed as shown in Fig. 54. 

(a) Find the moment 
of inertia of the section at 
tlie center. Ans. 252. 

(b) Find the moment 
of inertia of the section at J^ 
the distance x from the 
left end. Ans. 202.66. 

(c) What load P would 
break the beam at the center ? 




d-sl ^d-6" ^ 




^^p 

Fig. 54. 

Ans. 6,300 pounds. 

(d) How far from the left end should the hole, whose 

diameter is 8 inches, be bored, tha,t the beam may be as 

likely to break at this place as at the center, leaving out 

weight of the beam? Ans. 8.04 feet. 

Prob. 94. A heavy 10'^ I beam is supported at points 
20 feet apart, and overhangs a distance of 5 feet on each 
side. A load of 2,000 pounds is hung on each end, and a 
load of P pounds at the center. 

(a) What load P would break the beam ? 

Ans. 36,037 lbs. 

(b) What load P w^ould break the beam, supposing the 
end loads left off ? Ans. 34.037 lbs. 

(c) Supposing a 5 inch hole to be bored at the center, 
what load w^ould break the beam in each case ? 

Ans. 34,479 and 32,479 lbs. 

Prob. 95. What load applied at the center w^ould be 
likely to break a standard 6 inch w^rought iron pipe used 
as a simple beam of 20 feet span? Ans. 8193 lbs. 

Prob. 96. Compare the strength of two 2X10 planks 
placed one upon the other, with a single plank of the same 
wadth and 4 inches thick. Ans. as i to 2. 



98 

Prob. 97. Which is stronger when made of the same 
material and spanning an equal space, a 2X12, or a 4>^ X 
8'' beam? Assuming the beams to be 16 feet long, what 
would each cost at $20.00 per i,ogo feet board measure? 

Ans. 80 cents and $1.20 



Art. 



Beams of Uniform Strength. 



In order that a beam ma}^ be equall)^ strong throughout 

its length, the value of S must remain constant; or the 

unit stress at the outermost fibers must ever3^vvhere be the 

S I 
same. Hence in the formula M = , we may suppose 

the variable quantities to be M, I, and c. Thus, taking 
a cantilever wnth a load P at the end, and assuming the 
cross section to be everywhere a rectangle, the bending 
moment at any distance x from the free end is simply Px. 



For this and all other sections — 

c 



bh^ 



Hence the 



equation becomes 



Px = 



Sbh^- 




Fig. 55- 



If we suppose the width b to 

be constant, solving for h the 

6Px 
equation becomes h' =z 

From this, for any assumed value 
of X, the corresponding value 
of h can be found, giving an out- 
line similar to Fig. 55. 



99 



On tlie other hand, if we as- 
sume the depth h to remain 
the same, the varying width 
6Px 



b =z 



Sh' 



This will be 




Fig. 56. 



found to give a wedge shaped 
beam similar to Fig. 56. The 
section at the end however, 
and elsewhere, must be large 
enough to resist the shear. 

Any other assumed method of loading a cantilever, or a 
simple beam, will effect the value of the bending moment 
M, and the shape of the cross section may be changed 
as desired. Thus, taking the case of a simple beam uni- 
formly loaded and having everywhere a circular section, 



M = R X — 



wx 



and 



Ttd' 



2 c 32 

Hence the equation, by means of which the diameter can 
be found for any assumed value of x is 

;rd^ 



Rx — 



w x 



2 32 

Prob. 98. A wooden cantilever is to be 20 feet long, 
and to sustain a load of 4 tons at the end, with a factor of 
safetj^ of 6. 

(a) The cross section being everywhere a circle, what 
should be the diameter at the wall ? Ans. 23.5 inches. 

(b) Deduce the equation and find values for the diame- 
ters at points 5, 10 and 15 feet from the free end, that the 
beam may have equal strength at these sections. 

Ans. d^ = 54.3 x. d =: 14.8, 18.6, and 21.3 inches. 



CHAPTER VII. 

THE DEFEECTION OF BEAMvS. 

Art. 40. Equation of thk Elastic Curve. 

A loaded beam is always more or less bent, and some- 
times it is CI importance to calculate the amount of the 
bending, or the deflection. 

When uniform!}^ loaded or when concentrated loads are 
placed at the center of a span or at the end of a cantilever, 
this calculation is not difficult, but for other positions, or 
w^hen more than one load is applied it gencrallj^ becomes 
complex and unsatisfactor}^ In the following articles only 
such simple cases will be considered. The discussion, how- 
ever, involves a knowledge of elementary calculus, without 
which the student will be unable to do more than accept 
the results obtained on faith. 

The theory is based on the assumptions that (a) the 

stresses nowhere exceed the 
elastic limit, (b) the beam 
is bent so little that its hori- 
zontal projection is practi- 
cally the same as its length ; 
and (c) the cross section is 
either constant or varies defi- 
nately with the length of 
the beam. 

Let A be any point on the 
Fig. 57. neutral axis of the bent beam 




lOI 

whose co-ordinates, Fig. 57, are x and y, the origin being 
at O. Draw AR perpendicular to tlie neutral line ee and 
consider the elongation due to bending of a fiber whose 
length is ds. Let (p be the angle between AR and the 
axis of y and dcp h^ the increase in this angle correspond- 
ing to the length ds. 

Drawing DD parallel to AR, the elongation or compres- 
sion of the fibers whose original length was ds will be repre- 
sented by the shaded portion. Thus at the distance z from 
the neutral axis, this length has been increased by an 
amount K. But since the angle HBD is equal to dcp, the 
elongation K =z zdcp (a) 

If we designate by Sz the unit stress at the distance z, 

PI 

by Art. 9, the elongation due to this stress K = -t-— : = 

A Jb 

.q_1 ^ S ds 

-, or since the length considered is ds, K = — ^;t- (b) 



E ' ^ ' E 

or, from (a) and (b) za<p =1 — — or -— — = =r- (c) 

J-j Q.S J_/ Z 

Since the unit stress in beams is directly proportional to 
the distance of the fiber from the neutral axis. 



Sz z 



S c 



Sz S 



or = — , where S is the stress at the outermost 

z c 

fiber. Inserting this in equation (c) it becomes — i — = ti^ 

ds Ec 

Mc 
But, from the last chapter, the unit stress S = — ^ — 

XT d^ M ,,. 

Hence ^T = ET ' ^^^ 

Since BA and BC, Fig. 57, are perpendicular respec- 
tively to AR and OY, the angle ABC must equal the 



I02 



angle q) \ hence from the right triangle ABC, we have the 
relation, tan q) ---= — — . Since this angle is alwaj^s small, 



d 



dy 



tan qj will very nearl}^ equal arc </> ; or arc qj := — — , an.d 



by differentiation dqj 



d^ 
dx • 



For the same reason the 



arc ds A^er}^ nearly equals its projection dx. Making these 

dq^ _ d'y _ M 
ds~ "~ "cb? "~ ^ 



substitution's, equation (d) becomes 



or 



E I 



dx^' 



M 



[39] 



This is the fundamental equation by means of which the 
equation of the elastic curve and the deflection of beam.s may 
be determined. 



Art. 41. Cantilhvkrs. 




Let Fig. 58 represent a 
cantilever wuth a load P at 
the free end. Since it is im- 
K material where the origin of 
co-ordinates is taken, let us 
place it at O, and let x and 
■^D y be the co-ordinates of any 
point of the curve. 

Our fundamental equation 
d'y 



is HI 



dx^ 



M. But in 



P X, the moment being taken negative 



I03 
because it tends to produce a left-handed motion. Hence 

-12 

El-r-^ = M = — Px (a) Integrating twice 

dx 

El4^ = -_2^ -f c (b) 

dx 2 

Ely = -^ + Cx + C, (c) 

o 

where C and Ci are the constants of integration, and 

dv . 
whose values it is necessary to find, -z^^— is the tangent 

dx 

of the angle which the tangent line TT makes with a hori- 
zontal at the point x y, and if this line is itself horizontal 
at any known distance x from the origin, we can usually 

dy 
find the constant C by inserting: this value and Dlacmg - — 

" dx 

= o. Thus, in the present case, we know the tangent to 

the curve to be horizontal where the beam leaves the wall. 



12 



dy Pr 

or when x = 1. Hence K I— r^ = + C = o or 

dx 2 

PI' 
C = . In equation (c) the constant Ci can be found 

by inspection of the figure, for we see that when x = o 
y also = o; hence Q =: o. Equation (c), or the equa- 
tion of the elastic curve, becomes therefore, 

Ely = -^ + -^ [40] 

By assuming values of x, the corresponding values of y 
can be calculated, and the alastic curve constructed, if de- 
sired. Generally, however, it is enough to know the maxi- 
mum deflection AB. This can readily be found by making 
x = 1 in the above equation, by which we find, 

pr 
max. y = 

3E I 
15 



I04 




Let Fig. 59 represent a 

cantilever uniformh' loaded 

- - - -<A with w pound; 



[s per inch 

w X' 



(a) 



Fig. 59. 



/ E i-i— y = :si = 



EIv = 



w : 



24 



- + C X + C, 



(c) 



When X = 1, -r^- = o. Hence C = 



dx " ' 6 

When X 1= o, 3^ 1= o. Hence Cj = o 
The equation (c) of the elastic cur^-e therefore becomes, 



EIv = 



W X* W 1" X 

~24 ' 6 



[41] 



To find the greatest deflection AB, in equation 41 make 
X = 1. Hence, after reducing, 



max. V =. 



8EI 



Assume the beam to carry both a concentrated and a 
uniform load. As before, 

El4X = M = -" 
dx' 

dx 2 



•r X 



w X 



+ c 



E I y = - 

When X 
When X 



Px' 



24 



+ Cx + C- 



1, 



dy 
dx 



(a) 
(b) 
(c) 



o. Hence 



c = Il+^ 



O' y 



o. Hence Ci =: o 



I05 
Equation (c) of the elastic curve therefore becomes, 

Ely = 7 -7 — [42] 

6 24 2 6 

If in this we make x = 1 

V? + 3 w 1* 



max. y ■=. 



24 EI 

Prob. 99. A heavy 10'' I beam, 15 feet long, is used as 
a cantilever. What maximum unit stress would be caused 
by a load of 2 tons hung on the end, 

(a) leaving out of account the weight of the beam? 

Ans. 19,200 lbs. per sqr. inch. 

(b) considering also the weight of the beam ? 

Ans. 20,820 lbs. per sqr. inch. 

Prob. 100. What would be the greatest deflection caused 
by the above load in each case ? 

Ans. (a) 1.6 inches, (b) 1.7 inches. 

Prob. loi. What load at the end of a heavy 10^' I beam 
15 feet long, used as a cantilever, would stress it to the 
elastic limit, 

(a) leaving out of account the weight of the beam? 

Ans. 4,166 pounds. 

(b) considering also the weight of the beam? 

Ans. 3,829 pounds. 

Prob. 102. A load of 4 tons is hung at the end of a 
heavy 10'^ I beam, 10 feet long. What is the factor of 
safety? Ans. S = 26,320. f = 2.1. 

Prob. 103. Assuming the last answer given in Prob. 
loi, what would be the greatest deflection? 

Ans. 1.63 inches. 



io6 



Art. 42. Simple Beams. 








Consider a simple beam 
with a load at the center. 
p Taking the origin of co- 

^ ordinates (Fig. 60) at the 
P left hand support, as before 



A- 



Fig. 60. E I-p; = M 

dx" 

At any distance x between the end and center, 

^^^ = -^ = ^ (^) 

dv P X- 

^^ dT = -T' + ^ (''^ 

E I y = 4^- + C X + C, (c) 

In this case the tangent will be horizontal at the center, 
when X = — , -3 — ^ o. Hence C — 



2 dx 16 

When x = o, 3^ == o. Hence Ci = o 
The equation of the elastic curve therefore becomes 

Px' PPx r -, 

^^^=-—2 YT ^43] 

If in this we make x = 34 1, we shall get the greatest 
value of y. Substituting and reducing, 

pr 

the minus sign simply indicating that y should be meas- 
ured downwards. 

In a simple beam uniformly loaded, the origin being 
taken as before at the left-hand support, 



I07 



M = Rix 



^ W X' = ^ V/ 1 X 



>^wx^ Hence, 



d"'3^ w 1 X 



dx 

dx 



wlx^ 



wx 



wx 



+ C. 



Ely = 



w 1 x^ 

12 



1 



When X = — , 

2 



WX 
24 

dy 

dx 



+ C X + C, 



= o. Hence C 



-(a) 
...(b) 
-(c) 



24 



When X = o, y = o. Hence Ci = o 
Inserting these, the equation of the elastic curve becomes, 

w 1 x^ w X* w 1* X 



Ely = 



[44] 



12 24 24 

To get the greatest deflection, making x = >^1 we find 

5wl* 

max. y = ^ -^ ^ 

384 E I 

As before stated, it is immaterial where the origin of co- 
ordinates is taken, but for the sake of illustration let us 
assume it at the center of 
the beam, as in Fig, 62. 

^ _ 
dx'^ 



As before E I -^^ = M 




But at the point xy, 

T. M \ w/1 

M = Ri X 1 — 

\ 2 / 2 V2 

d^3^ w 1^ w x^ 

^ ^ d? - ~8 T" 



Fig. 62. 



wV 



wx 



X = 



EI 



dx 



wTx 



Ely = 



wFx^ 

16 



wx 

6 
4 



W X 



24 



+ C 



+ C X + C, 



(a) 
(b) 
(c) 



loS 

In this case, the tangent being horizontal at the center, 

when X =^ o, -~~ = o. Hence C = o 
dx 

when X = o, 3- = o. Hence d == o 

The equation of the elastic curve is therefore, 

122 4 

-^ ^ W 1 X W X 

E I y = — — 

16 24 

The greatest value of y will occur vrhen x = }^1, or 

, 5 w r 

max. y = -r ^ -^s -. 
384 HI 

which is the same value as that obtained above, excepting 
that the plus sign indicates an upward measurement. 

Prob. 104. A yellow^ pine beam, 20 feet long, 8 inches 
wdde, and 12 inches deep, is supported freely at each end 
and loaded uniformly until the greatest unit stress is 2,000 
lbs. per sqr. inch. 

(a) Find the total load. Ans. 12,800 pounds. 

(b) Assuming the answer to (a), find the total deflec- 
tion. Ans. 1.7 inches. 

Prob. 105. What I beam should be chosen to sustain 
a concentrated load of six tons at the center, the span be- 
ing 12 feet, and the factor of safety 6? Ans, Light 12''. 

Prob. ic6. What deflection would the above load cause ? 
(Prob. 105.) Ans. 0.104 inch. 

Prob. 107. A light 8'^ I beam having a span of 20 feet, 
is deflected one inch by a uniform load. 

(a) Find the total load. Ans. 10,100 pounds. 

(b) What is the greatest unit stress caused by it ? 

Ans. 17,314 pounds. 



log 



Art. 43. Rkstrx^inbd Bkams. 



Let us now consider a beam 
fixed at both ends, and with a 
load of P pounds at the center. 



As before B I- 



dx^ 



= M 




In this case however the bend- 
ing moment can not be directly 
stated on account of the unknown bending ilioment where the 
beam enters the wall. Taking the origin of co-ordinates 
at the left end, and calling the moment and vertical shear 
at this point respectively M' and V, we may make use of 
the general expression for M derived in Art. 33 ; 

M = M' + V X - -^^ 

2 

Since there are no loads to the left of the support, the 
vertical shear at that point must equal the reaction, and 



since there is no uniform load >^wx^ 



o. Hence 



EI 



HI 



dV 

dx' 
dx 



Ely = 



= M'x + 
:: "1 



Px 



Px^ 

4 

P_x^ 
12 



+ c 



+ Cx + C, 



(a) 
(b) 
(c) 



We have here three unknown constants to determine, M' 
C and Ci ; but, by inspection of the figure, we see that the 
tangent will be horizontal both at the wall and at the cen- 

dy 



ter. Hence when x = o. 



dx 



o, and C = o 



no 

when X = >41 ~- = o, and M' ==^ 

dx 8 

when X == o, y ^^ o, and Ci = o. 

Inserting in (c), the equation of the elastic curve becomes 

P 1 x^ P x^ 

The value of y will evidently be greatest when x = >41. 

Hence max, y = — -— 

192 K I 

Having found the bending moment at the support, that 

at any other point may readily be determined by means 

PI P ^ 

of the equation, M = M' + V'x == + 



2 

The general expression for vertical shear is, see Art. 33, 
V = V' — wx — P, but in this case there are no loads 
between the center and left support. Hence the vertical 
shear for all points V = V' = >^ P Just to the right of 
the center, V = >^P — P == — >^P. The shear therefore 
passes through zero at the center, which must be a danger- 
ous section. Inserting x = >41 in the equation above, 

PI PI PI 

^■^ = -— + — = +^ 

The bending moment at the center is therefore the same 
as at the wall, the only difference being in the sign. Be- 
tween the wall and center, the curvature changes from 
convex to concave. There must be then a point where the 
beam is not bent either way, or where the bending moment 
is zero. To find this point of inflection, make the general 

PI P x 1 

expression M =■ — -h =■ o Hence x = — 

or the inflection point is midway between the wall and 
center. 



Ill 



lyCt Fig. 64 represent a beam /j 
fixed at both ends, and uni- 
formly loaded with w pounds 
per inch of length. As before 




M -- 


= M 


' -f V 


r 

x — 


EI 


d-^y 
dx' 


= M' 


+ 


EI 


dy 
dx 


= M'x -f 


KT 


■\r 


M'x' 


_J_ 



Fig. 64. 



w 1 X 

2 

W 1 X^ 

4 
wl X^ 



WX 



W X 



W X 



+ c. 



(a) 
(b) 



12 



^4 



+ Cx + C, (c) 



When x = o, 



d^ 
dx 



When X = — , -^ 
2 dx 



o. Hence C = o. 



o. Hence M' = 



wV 



12 



When X = o, y = o. Hence Ci = o. 

Inserting these values in (c), the equation of the elastic 
curve becomes, 



Ely 



wPx^ wlx^ 



W X 



[46] 



24 12 24 

Making x = 7^1, the greatest value of y will be found 

wl' 
max. y =. 



to be 



384 B I 

The greatest bending moment between the supports will 
be at the center where 

wl' wl' 



M = 



w P . w 1' 



+ 



12 • 4 8 24 . 

This being less than the negative moment at the ends, the 
beam would most likely break where it joins the wall. 
16 



112 



Tlie inflection points may be found by making 



IM 



W P W 1 X 



Vv^ X 



12 • 2 2 

From which, solving as a quadratic, 
X = 1 



:= O 



I 
> 12 

Let us now consider a beam 
fixed at one end and support- 
ed at the other and uniformly 
loaded. The support being on 
the same level, as before, 
Fig. 65. M = M' -^ V'x — >^wx' 

but since there are no loads to the left of the free end, 
M' = o, and V' = R. Hence 




^ dx^ - -^^^^ 



W X 



El4^ = 
dx 



Rx- 



2 

W li 



H- C 



Ely 



Rx^ 



w X 



+ Cx + C, 



(a) 
(b) 
■(c) 



o 24 

In these equations there are three unknown constants, R, 
C, and Ci , but 



when X = 1, 



dy 



:= o. Hence C 



RP 



wP 



dx 2 

when X = o, y = o. Hence Ci = o. 
when X = 1, 3^ = o. Hence from, (c) R = §^wl. 

The equation of the elastic curve is therefore, 

13 4 13 

:iL w X w 1 X 

±v iy — 



[47] 



48 24 48 

The maximum deflection cannot be found directl}^ from 



113 

this equation, for the reason that it is not apparent where 
the tangent to the elastic curve betvreen supports is hori- 
zontal. This tangent point can, however, be found from 
equation (b) by placing it equal to zero and solving for x; 
first inserting the values of R and C just found. Thus, 

^ ^ dy 3 w 1 x'^ w x^ w P 

K I — r^ = - — : — TT- = o or 

dx i6 6 48 

— 8x' -f 9ix' = r 

This equation cannot readily be solved algebraically, 

and should be treated graphically on plotting paper. One 

root will be found to be, x = 0.42 1, which inserted in 

w 1^ 
equation (47) gives max. y = 0.0054-—— 

The inflection points may be found by placing equation 
(a) equal to zero, and solving for x. Thus 

-^ 3 v/ 1 X wx^ 3 , 

M = -^—- =0 or X = -^1 

82 4 

From equation (a) the bending moment at the wall is 

wl wl 

M = -^^— = — 

82 8 

The maximum bending 'moment between the supports is 
determined as demonstrated in Art. 32, by finding where 
the vertical shear becomes zero. Thus 

V = R — wx = ^wl — wx =■ o 

From this we see that M is greatest where x =: s/qI. In- 
serting this value in the expression for M, we see that 

9 v/ 1" Q w 1^ , Q w 1^ 

max. M = -^- — ^ =z -f ^ 



64 128 128 

The beam is therefore most likely to break where it enters 
the wall. 



114 

The student who wishes to pursue this subject further 
will find it admirabh' treated in Merriman's Mechanics of 
Materials, Chapter IV. 

Prob. loS. A light 8 inch I beam fixed at both ends and 
having a span of 20 ft., is deflected one fifth of an inch b}- 
a uniform load. 

(a) Find the total load. Ans. 10,100 lbs. 

(b) What is the greatest unit stress caused by it ? 

Ans. 11,540 lbs. per sqr. inch. 

(c) Find the inflection points. 

Ans. 15 ft. 9.6 inches, and 4 ft. 2.4 inches. 

Prob. 109. A pulley weighing 200 lbs. is fastened to a 
2>2 inch soft steel shaft midwa}' between the hangers, the 
distance between the hangers being 10 ft. Find the great- 
est unit stress caused b}- the weight of the shaft and pulle}- 
and the maximum deflection, considering the shaft as fixed 
at both ends. Ans. S = 1.955 max. y = 0.032 inch. 

Prob. no. A cast iron pipe, whose diameters are respec- 
tively 8 and 10 inches, is of such a length that it is stressed 
up to the elastic limit b}* its own v\'eight, one end being 
fixed and the other supported. 

(a) Find the length in feet. Ans. 126.5 f^^^- 

(b) Find the deflection in inches. Ans. 19.2 inches. 



CHAPTER VIII. 



STRESSES IN, AND STRENGTH OF SHAFTS. 



. Aet. 44. Thk Twisting Moment. 

Let figure 66 represent the end of a shaft twisted by a 
force of P pounds acting at the end of a lever-arm whose 
length is 1 inches, and at right angles to it, the point O 
being the axis of rotation. 

The product of this force and its 
lever-arm, or PI, is known as the 
twisti7ig moTnent. This must be 
resisted by the material. Upon ap- 
plication of the force P, the arm 
originally occupjdng the position 
OA moves through a small angle to 
OB, in which position we will as- 
sume that the moments of the stresses 
in the shaft exactly balance the external twisting moment 
causing them. The fibers most distant from the point O 
are most distorted. At O no distortion occurs, and evidently 
the amount of distortion will be directly proportional to the 
distance of the fibers from the axis of rotation O. This 
point usually coijicides with the center of gravity of the 
section. 

A line originally passing through a, and parallel to the 
axis O becomes approximately a spiral; and if we suppose 




Fig. 66. 



ii6 

one end of the shaft to be fixed, the angle A O B will be 
proportional to the distance of the point of application of 
the external force P from the fixed end. Thus if this force 
is applied at twice the distance, the angle becomes twice as 
great; at half the distance, half as great, etc. The distor- 
tion of any particle with respect to one adjoining is not 
however, changed. Hence while the total twist of a shaft 
depends upon its length, its strength does not; although, 
as wall be seen later, the bending action due to the w^eight 
of the shaft itself, and of the pulleys and gearing' attached 
to it, and stresses caused by belts, etc., materially modifies 
this law. 

Prob. III. A weight of i,ooo lbs. is hung on a crank- 
pin, the crank radius being 12 inches. Find the twisting 
moment when the crank makes angles of 0°, 30°, 45°, 60°, 
and 90° with the vertical. 

Ans. o, 6, coo, 8,484, 10,392, 12,000 lbs. inches. 

Art. 45. The Resisting Moment. 

Since the distortion produced in a shaft is directly pro- 
portional to the distance of the fiber from the axis of rota- 
tion, it follows from law A, Art. 7, that the stress is also 
proportional. Thus if S is the unit stress at the outer- 
most fiber whose distance is c inches from O, the unit 
stress at any distance z may be found from the proportion 

— ^ = — or Sz == — lbs. per sqr. inch. 

S c c "" 

If the area of this fiber is a square inches, the total stress 

S 7 a 
acting on it is — ^— pounds. The moment of this stress 



117 

*^ z a 

is evidently — X z pounds inches. Similarly the re- 

c 

sisting moment of any other fiber whose distance from O is 

Zi inches, and area ai square inches, is -^^^—. The total 

c 

resisting moment is therefore, 

So 

— (az^ -f- aiZj^ ~f- as z/ -{- etc.) lbs. inches. 

The sum of the series within the brackets is known as 
the polar nioinent of inertia of the section. Designating 

this by the letter J, 

S T 
Resisting moment =: ^^ 

Since the resisting moment must exactly equal the twist- 
ing moment causing it, we have the fundamental equation 
for the strength of shafts, 

p 1 = ^ [48] 

By means of this equation most of the problems arising 
in practice can be solved. 

The stresses caused by twisting a shaft are closely allied 
to shearing. One section tends to shear away from the 
adjoining, and it is customary to assume the values of Ss 
and Cs given in Tables IV and V. 

Prob. 112. In a steam engine in which the stroke is 3 ft. 
there is a constant pressure of 10,000 lbs. upon the piston. 
Find the twisting moments on the shaft when the crank 
makes angles of 0°, 45°, and 90° with the direction of the 
piston-rod, the length of the connecting-rod being 10 feet, 
and leaving out of account the weight of the moving parts. 

Ans. o, 113,720, 120,000 lbs. inches. 



iiS 



Prob. 113. Upon a lever-arm inclined 6c° to the horizon- 
tal, are hung weights of :^o, 40, and 50 lbs. at distances 
from the axis of rotation of 3, 4, and 5 feet respectively. 
Find the resultant twisting moment. 

Ans. ^,coo lbs. inches. 



Art. 46. Polar Moment of Inertia. 

The polar moment of inertia may be determined directly 
by the use of the Calculus, or b}?- the aid of the following 
proposition- 
In figure 67 let O be the center of rotation about which 

we wish to find the moment of 
inertia J. Through O draw 
two lines at right angles to 
each other, and let the distance 
of any elementary area a from 
these axes and O be designated 
by X, y, and z. The moment 
of inertia of this area about the 
axis YY is ax", about axis XX 
Fig. 67. is ay^ and about the point O is 

az^. From the right triangles in the figure, x^ + y^ = z^. 
Multiplying through by a we have ax^ + ^7^ ^^ ^^^- ^^^ 
any other area ai v/hose distances are Xi, yi, and Zi we have 
ajXi^ + ^i^V = ^i^i^- Adding these equations, 

(ax' + a^xi' -f aox/ + etc.) -{- (ay' + a^yi' + a.y,' + 
etc.) = ( az' + aiZi' + aaZa' + etc.) 

But the sum of the quantities within the first brackets is 
the moment of inertia of the figure about the axis YY; that 
within the second is the moment about the axis XX; v/hile 




119 

the other side of the equation represents the moment of 
inertia about the point O, or the polar moment. Hence we 
see that the "luomcnt of inei^tia of a section about the axis of rota- 
tio7i is equal to the sum of the niotneiits of inei'tia of this section 
about two axes at right angles to each othc/'- passing through 
the axis and lying in the plane of the figure^ or expressed in 
S3'mbols, 

J = I -1- ii 

Thus, let the figure be a circle whose diameter is d, the 
point O being at the center. From Art. 36, the moment 

TTd* 

of inertia about the axis YY is —- — ; about the axis XX 

64 

it has the same value. Hence 
ror a circle, J — 



64 64 32 

Similarly it may be proved that. 

For a square whose side is d, J ^ 



d^ 



6 

For a rectangle with sides b and d, J ==: + 

^ -^ 12 ' 12 

The distance c, from the axis of rotation to the most 
remote point of a section, may readily be seen to be 

For a circle c = — 

2 

For a square c = d-*/^ 



For a rectangle c = — -\/ b^ -j- d^ 

Prob. 114. Find the values of J and c for a rectangle 
the length of vv^hose sides are 4 and 5 inches. 

Ans. J = 73.42 c = 3.2. 
17 



I20 

Prob. 115. Find the values for J and c for a lieavj^ 10'' 
I beam. Ans. J = 198.3 c = 5-54- 

Prob. 116. Find an expression for tlie values of J and 
c for a hollow shaft whose diameters are d and di. 

Prob. 117. Compare the strength of a shaft whose diame- 
ter is 6 inches with that of a square shaft, the length of 
whose side is also 6 inches. Ans. As 1.2 is to i. 

Prob. 118. What force at the end of a lever-arm of 10 
feet would twist apart a soft steel shaft having a diameter 
of 2 inches? Ans. 746 pounds. 

Prob. 119. What should be the diameter of an iron shaft 
to sustain a force of 10,000 lbs. acting at the end of a lever- 
arrj. of 10 ft. the factor of safet}- being 6? Ans. 5. 11 inches. 



Art. 47. Transmission of Power. 

Let a shaft, acted upon b}' a force of P pounds at the 
end of a lever-arm of 1 inches, be turning at the rate of N 
revolutions per minute. During one turn P will move 

2 TT 1 

throuo^h a distance of 2 ^rl inches, or feet, and the 

12 

2 TT 1 P . . 

work done will be foot pounds. In one minute it 

12 

2 TT 1 P N 

must be foot pounds. If this is divided by 

12 

33,000, we shall have the horse power transmitted. Call- 
ing the horse power H, 

27rlPN PIN, ,, 

H = = (nearly) 

12X33.000 63,000 



121 



or P 1 = — ^^-^ — — Substituting this in equation 48 

S J 63,000 H , , . - 

^ — ^ - from which 



c N 

H = —^ [49] 

63,000 c 

For a round shaft — = : — ~ ==■ - — ~- Substi- 

c 32 

tuting above and reducing, 

SNd 
321,000 ^ \ SN 



d 




Ttd' 


2 




16 


3 / 


H 


- 



H = or d = 6S.5^^^ :. [50] 



For a square shaft whose side is d inches, it may be 
similarly shown that, 



H = -— or d = 64.5-4/. L51J 

267,000 ^^VSN 

In using the above equations a proper factor of safety 
should be chosen. It is customary to assume a working 
strength of from 8,000 to 9,000 lbs. per sqr. inch for iron, 
and from 10,000 to 13,000 lbs. per sqr. inch for steel, im- 
phdng factors of safety lying between 5 and 6. These are 
great enough for ordinar}^ shafting, but in cases where the 
twisting moment is liable to a sudden increase, as for ex- 
ample in rolling mills, a much greater factor must be em- 
ployed. For the crank-shaft of an engine it should at 
least equal 10. 

Prob. 120. A soft steel shaft making 150 R.P.M. is to 
transmit 500 H. P. with a factor of safety of 10. What 
should be its diameter? Ans. 5.73 inches. 

Prob. 121. How many horse power would a soft steel 



122 

shaft making 150 R.P.M. transmit, the diameter being 6 
inches and the factor of safety 10 ? Ans. 575. 

Prob. 122. An 18'' X 30'' steam engine is to have an 
average steam pressure of 40 lbs. per sqr. inch, and to 
make 80 R.P.M. with a factor of safety of 15. What is the 
H.P. of the engine, and what should be the diameter of 
the soft steel crank-shaft? Ans. PI = 123.4 <^ "^ 5 inches. 

Prob. 123. What should be the diameter of a soft steel 
shaft for a steam turbine making 10,000 R.P.M. in order to 
transmit 25 H.P. with a factor of safety 10 ? 

Ans. 0.52 inches. 

Prob. 124. Assuming the data of the last problem, what 
should be the diameter w^ith a speed of 100 R.P.M. ? 

Ans. 2.41 inches. 

Prob. 125. A belt passing over a flywheel transmits 
200 H.P., the diameter and speed of the wheel being 20 feet 
and 75 R.P.M. What must be the difference in the ten- 
sion between the driving and slack sides of the belt ? 

Ans. 1,400 pounds. 

Prob. 126. * Jones and Laughlins give the formulce, 



d 



\ N \ N 



the first being for ordinary turned wrought iron shafts, 
and the second for cold rolled wrought iron shafts. What 
working unit stresses do these imply ? 

Ans. 5,140 and 8,560 lbs. per sqr. inch. 

* Merriman's, "Mechanics of Materials," Chap. VI. 



CHAPTER IX. 
COMBINED STRESSES. 

Art. 48. Long Columns. 

When the length of a column does not exceed ten times 
its least transverse diameter, its strength may be estim^ated 
by the formula P = A Sc , when P is the load in pounds 
applied at the end in the direction of the length, A the 
cross section of the column in square inches, and Sc the 
unit stress caused by the load. When, however, the length 
is greater in proportion to diameter, this formula cannot be 
safely used, because in addition to direct compression, an 
additional stress is produced by bending of the columu. 

Thus let Fig. 68 represent a long column DB, 
with a load of P pounds upon the upper end. 

If the beam did not bend at all, the unit stress pv' 

P ' 

caused by this load would be Si = — r- : but if ' 

A I 

bent as shown there will be an increase due to | 

flexure, which ma}^ be computed by the formula 

M c , ^ 

S2 = - — ^ — as explained in Art. 35, where M 'I 



is the greatest bending moment, I the least mo- I 

ment of inertia about the neutral axis at the sec- [ 

tion under consideration, and c the distance to B^ 
the most remote fiber from the axis. The total 
stress at this section is therefore, Fig. 68. 

P Mc 

s = s. + s. = ^ + ^ 



124 



The bending moment is evidently greatest where the 
beam is most bent, and where we will call the deflection 
m, and its value (see the Fig.) is M = P m. The mo- 
ment of inertia I may be expressed in terms of the area 
A and radius of g3Tation r of the cross section, or I =: 
Ar'. Substituting above, 



«= A + 



P mc 



I + 



m c 



Ar A\ • r 

The deflection m depends upon the manner in which 
the column is held at the ends. It is also found to vary 
directly as the square of the length of the column, and in- 
versely as the distance c. 

Thus in Fig. 69, (a) rep- 
resents a column free at both 
ends, bending like a bow. 

In (b) one end is fixed 
vertically, or in the direc- 
tion of the force P, and there 
is a double curvature. 

In (c) both ends are fixed 
similarly, and produce a tri- 
ple curvature. 

The deflection m is there- 
fore equal to some quantity 
which we will call k, mul- 
tiplied by 1' and divided by 



Fig. 69. 
kl^ 



c, or m 



Substituting this in the last equation, 



[52] 



S = -T-1 I + T i 

A\ T J 

This is known as Rankine's or Gordon's formula for the 



125 

strength of long columns. It is to a large degree experi- 
mental and values of k vary between quite wide limits. 
For rough calculation the follovvdng table, taken from Mer- 
riman's " Mechanics of Materials," may be used, but for 
important work, especially when the column or " strut " is 
*' built up " or made of several pieces, as for example in 
bridge work, resource should be had to books giving 
especial attention to the subject, or to trade catalogues and 
hand-books. 



TABLE XIV. 
Vai^uks of k. 

Material. Both ends Fixed. Fixed aud Round. Both ends Round. 

Timhpr 1 1-78 4 

1 imoer 3 qj) q 30 3000 

Ca^f Irn-n 1 1-78 4 

\^ast iron 5 q q q 5 00 5 000 



Wrought Iron .3-g-J- 



1-78 4 



3 6 3 6 

e^-ppi 1 • 1-78 . 4 

^'^^^^ 2 5 2 5 2 5 

A column with a fiat end, against which the load rests, 
may be considered as fixed ; when pinned or held b}^ a pin 
passing through the axis, it is treated as though round. 



Art. 49. Radius of Gyration. 

The radius of gyration can readil}'^ be found by aid of 
the equations given in Art. 36. Thus, 

I = A r^ or r' = -^ 

A 



12 


+ dr 


d- 4 


i6 

- dr 



126 

For a circle I ::= -— — and A -= Hence r* = 

04 4 16 

d' 

For a square of side d, r" =z 

12 

For a rectangle of least side d, r^ — 

For a hollow circular section, r' = 

For a hollow square section, r = 

12 

In a similar manner the radius of gyration of a wrought 

iron I or deck beam may be obtained, using the values 

given in Table XII. 

Prob. 127. P'^Ind the least radius of gyration of a heavy 
12'"' I beam. Ans. r = 0.86. 

Prob. 128. Find the least radius of gyration of a light 9'' 
deck beam. Ans. r' = 0.35. 

Prob. 129. Find an expression for the radius of gyration 
of an equilateral triangle about an axis passing through 
the center of gravit)^ and parallel to a side. 

h'^ b'^ 

18 24 

Prob, 130. A hollow column is made of i^' boards, each 
side being 6'' wide. Find the least radius of g3^ration. 

Ans. r = 4.333- 

Art. 50. Practical Applications. 

As an illustration of the use of Rankine's formula, let it 
be required to find the diameter of a soft steel piston rod in 



127 

an i8'' X 30'' steam engine wliicli is to sustain a steam 
pressure of 125 lbs. per sqr. inch, with a factor of safety-- of 15. 
The area of the piston is 254.5 ^1^- ii^ches and the total 
pressure acting upon it, P = 254.5 X 125 = 52,312 lbs. 
The s:reatest allowable unit stress in the rod S == 80,000 
-f- 15 = 5>330 lbs. per sqr. inch. Since the rod passes 
through a stuffing box at one end, and is fastened to the 
cross-head at the other, it may be considered as fixed at both 
ends, and from the table k = i ^- 25,000. Its length may 
be taken as somewhat greater than the stroke, or sa}^ 36 

TT d^ 

inches. The cross section of the rod A = and the 

d^ 
square of the radius of <ryration r^ = ~. Rankine's for- 

10 

mula may therefore be written 

4 P / , i6kr \ _ 4P(d^ + 16 kl^) 
^ - '^c^V "^" d^ / ~ TZd' 

or ;r S d' — 4 P d' = 64 P k V 

,4 4P^2 64 p k r- 

01: 6r — —^ d' = -^ — - — 

77" b TT S 

Substituting in this the values given, and reducing 
d'— 7.72 d' = 6.4 

Solving this as a quadratic d^ =: 8.46 and finally d r=r 2.9 
inches, vv^hich is the diameter required. 

Let it be required to find the thickness of a hollow cast 
iron column with flat ends whose outside diameter is 10 
inches, in order that it ma3^ sustain a load of 80 tons ; its 
length being 20 feet and the factor of safety 10. Rankine's 
formula becomes 

^ _ 4P / , i6kr \ _ 4P(d^-f d,^+ lokf) 

~ 7r{d' — d,^)i/ "^ d^+ diV "" 7r(d^ — d,^) 

14 1 4^12 J4 4 ^ J2 64 P 1 12 

18 



128 



Substituting given values and reducing 



di^ 4" 20.4 di' =^ 10,000 — 2,040 — 3,686 = 4,274 
Solving as a quadratic and using the plus value, the inside 
diameter is found to be d = 7.5 inches. The column should 
therefore be about ij^ inches thick. 

Prob. 131. What should be the diameter of a soft steel 
piston-rod in a 36'' X 48'' steam engine, the steam pressure 
being 200 lbs. per sqr. inch, the length 56 inches, and the 
factor of saiet}^ 15 ? Ans. d =: 7.1 inches. 

Prob. 132. What load would crush a heavy 12" I beam 
10 feet long with fixed ends? Ans. 528,300 pounds. 

Prob. 135. What should be the thickness of a square 
3'ellow pine post 20 feet long, with fixed ends, to sustain a 
load of 20 tons, the factor of safety being 5 ? 

Ans. d r=: 9.7 inches. 

Prob. 134. What load would crush a 10''' steam pipe 
used as a flat ended post, the length being 30 feet ? 

Ans. 404,200 pounds. 

Prob. 135. The internal diameter of a cast iron column 
with fiat ends, being "jVi" , what should be its external di- 
ameter under a load of 80 tons, the length being 20 ft., and 
the factor of safetv 10? Ans. 10 inches. 



Art. 51. Compound Columns. 

A column is frequently made of two I beams joined to- 
gether by lattice bracing as shown in Fig. 70. The moment 



129 




of inertia of the bracing is iisuall}^ neg- 
lected, and the I beams should be placed 
at such a distance apart that the moment 
of inertia of the beams about the axis XX 
(see Fig. 70) is the same as that about 
YY. 

Thus let us assume 10'' I beams. The 
moment of the two beams about YY is 
(see Table XII) 2 X 187 = 374. In order 
to get the moment about XX, we must 
make use of the formula I n: I' — A z^ or 
I' = I + A z' (see Art. 36) . The moment 
of the left-hand beam about axis ab is I 
= Ii = 1 1.3. The area of the beam is 
13.5 sqr. inches, and the proper distance Fig. 70. 

between ab and XX is the unknown quantity we wdsh to 
find. Hence the moment of inertia of the two I beams about 
the axis XX is 

1 = 2 (11. 3 + 13.5 z') 

Placing this equal to the moment about YY already de- 
termined, we have for our equation 

22.6 -|~ 27 z^ = 374 or z =z 3.6 inches. 

The width of the flange being greater than this the beams 
should touch each other. If now we apply Rankine's for- 
mula to this compound beam, it is only necessary to find r* 
and to choose a proper value for k. 



Thus 






I _ 374 



A 13-5 

ends fixed, roughly k =z g-g-J-oio 



27.7 and, supposing both 



i-;o 



Let it be required to build a column of tv.'O 2 X lo wooden 

beams with side coverings of one 
inch boards as in Fig. 71. What 
should be the distance from center 
to center of the beams ? The mo- 
ments of inertia about the axes XX 
and YY should be equal. The mo- 
ment of inertia of the two beams 



Y 



1^ 


iX 




-•,-, _ H 1- 


1 

I 
1 

_ L 

1 
1 
1 
1 


_ i_ - 

1 
1 
1 
1 


2" 


1 > 


b 


;x 





Y 



about YY is 



2 X 2 X (10)" 



12 



000 



Fig. 71 



AVe see that about the axis c d 



To this must be added the moments 
of the two covers. Making use 
again of the formula (58) 
r =: I -f A z' 

(2S —2) X i' 



I = 



12 



A=(2s^2) X I sqr. inches and z = ^y^ inches. 
Hence for one beam 
2 s -(- 2 



r z= 



(2S + 2) 



V 



p. 25 = 60.67 (s — I) 



The total amount about YY is therefore 
333 -h 2 X 60.67 (s + i) = 454-3 -f- 121. 3 s (a) 

Let us next find the moments of the two beams about XX. 

10X2'' 80 



xYbout axis a b I = 



A = 2X 10 = 20 



12 12 

sqr. inches, and z = s. Hence, for both beams, 

80 



1 = 2 



V 12 



+ 20 S' 



To this must be added the moments of the covers about 
iX(2S + 2)' _ 8 (s -f i)' 



XX, or 2 X 



12 



I.^I 



The total moment of inertia about XX is therefore, 

80 . A . S_ 

~6 



^ + 20SM + ^(S -h I)^ 



(b) 



Placing the total moments about XX and YY equal, 

80 
^- -^ 20s- I -t -^ 



454.3 + 121.3 s =z 



+ 20SM + -—(S + l)^ 



Cubing (s 4" i)' combining and arranging terms, and 
omitting unimportant decimals, the equation becomes 

s' + 33 s' — 88 s = 330 

This equation of the third degree can not readily be solved 
algebraically, but as in nearly all equations the roots of 
practical importance can be quickly found by the aid of 
plotting paper. Thus let us call the right-hand side of the 
equation x, and con- 
sider X to vary with 
s. Draw axes XX, 
YY at right angles 
to each other. Let 
us now give some 
value to s, as say 2''. 
Then x would equal 

2=^ -f- 33 X 2^ — 88 X 
2 = ■ — 36. This is 
so much less than 
330 that it would evi- 
dently be better to riG. 72. 
assume a larger value of s, say s = 3''. In this case x 
equals -|- 60. Lay off then s = o a = 3 inches and assum- 
ing any convenient scale, make the corresponding value of 
X = o b = 60, say 0.6 of an inch, draw the dotted lines and 



5 
4 


Y 


5^ 


-^ 


— 


a 


P^ 


1 1 






2 










1 










X 


lb 


!b, Ib3 


|b. 


X 





V ' ^^ 


1 
- - ^ 







132 

find one point (P) of the curve which will show hovv* x 
varies with s. Let s now equal 4 and 5 inches. The cor- 
responding values of x are 240 and 510. Using the same 
scale and in the same way, find Pi and P, and through 
these points draw the curve, three points usually being 
sufficient, provided one value of s gives a greater value of 
X than that in the equation to be solved. Make now x 
equal to 330 and la3dng off o b., equal to this number, erect 
the perpendicular b P3. It will cut the curve at Pg and the 
distance bg Po multiplied by the scale adopted will give us 
the value s which will satisfy the equation. By this method 
it will be found that the distance from center to center of 
the beams forming the column should be 2 s = 2 X 4.38 = 
8.76 inches. Tlie total moment of inertia of the column 
from either equation (a) or (b) is 

454'3 -r 121. 3 X 4-38 — 985-^ 

The area of the cross section is 

2(2 X 10 + I X 10.76) =61.5 sqr. inches, and 

,. _ i_ ^ 985-6 ^ ^^ 
A 61.5 

Assuming now a proper value for k, say for fixed 

° x- i- 3,000 

endS; the strength of a column built in the way indicated 
can be determined by the use of Rankine's formula. 

For example, assuming the material to be yellow pine, 
and the factor of safety 10, such a column 20 feet long 
would sustain a load of 

^ AS 6 2 . ^ /\ 7 -^u , 

P r= =z ^ -^ — — =1 20,400 pounds. 



AS 


62.5 X 730 


kl' ~ 


(20 X 12)' 
3,000 X 16 



oo 



Prob. 136. What should be the distance from center to 
center of two light 8^' I beams used as a column when joined 
b}^ lattice bracing? Ans. 6.24 inches. 

Prob. 137. Two wooden beams each 2 X 10 inches are 
placed 10''' apart from outside to outside. What should be 
the thickness of covering boards? Ans. 0.77 inches. 



Art. 52. Combined Tension and Fi^exure. 

Let Fig. 73 represent a beam subjected to loads P2 tend- 
ing to bend it, and also to 

others Pi acting in the di- ^ ^ 1 1 ^7^ 

rection of its lensfth. The A I A 



"O 



B 



a 



former produce tensile and 

compressive stresses which Fig. 73. 

have been discussed in Chap. VI ; the latter an additional 

tensile stress which ma}^ be calculated by the formula vS, = 

Pi -^ A. The total unit tensile stress is therefore, 

S = Si -f- S, :^ ^^ + ^' 



I ' A 

Thus let the above Fig. represent a light 10'' I beam of 
20 feet span, loaded at the center wdth a weight of 4 tons; 
and also subjected to a pull of 10 tons lengthwise. The 
greatest bending moment is at the center and is Pol -^- 4, 

8,000 X 20 X 12 It. • 1 

or M = z=z 480,000 lbs. inches. 

4 

From Table XII, I ^- c ^= 30. Hence 

MC 480,000 r AU • T 

bi =. — =r— =. -. =1 16,000 lbs. per sqr. inch. 

I 30 

The total stress lengthwise. Pi z=z 20,000 lbs.; the cross 



section of the beam, (see Table XII), is 9 sqr. inches, and 
the unit tensile stress, 

Px 20,000 . 

oo = —r— r=: z== 2,222 lbs. per sqr. inch. 

A 9 X- 1 

The total unit tensile stress is therefore, 

S =: Si + S2 = iS,222 lbs. per sor. inch. 

Prob. 138. What should be the diameter of an iron bar 
10 ft. long stretched by a force of 50 tons, and subjected at 
the same time to a transverse load of 10 tons applied at the 
center, choosing 5 as a factor of safety and assuming the 
ends to be fixed as in Art. 43 ? Ans. d =: 7 inches. 



Art. ^:^. CoMBixED Tension or Compression and 

Shear. 

The discussion of the case where material is subjected to 
tension or compression and also to a shearing force acting 
at right angles, is somewhat complex and belongs to a more 
advanced work on the subject. The resultant stresses 
caused b}^ these forces ma}-, however, be expressed by the 
equations 

For resultant tensile stress. 



Sf / ^ 



For resultant shearing stress, 



S' 



s's = ys% -h -^ [54] 



135 



In which S\ and S's are the greatest unit stresses due to 
the tensile and shearing forces at the section under consid- 
eration, the same equations being used in case a compres- 
sive force is employed. Thus, a^ in Fig. 74, let an eye-bolt 
be subjected to a shearing 

L 



force of 4,000 lbs. and also 
to a tensile force of 5,000 
lbs., the diameter of the 
bolt being one inch. The 
unit shearing stress due 
to the 4,000 lbs., will be 

_ 4 X 4>OQO , ,,0 

^^- 2X^d^ ^ ''^^^ 
lbs. per sqr. inch. The 
unit tensile stress due to 
the 5,000 lbs. will be 

_ 4 X 5)QQQ _ 




4000 



St 



Ttd' 



5000 

Fig. 74. 

6,370 lbs. per sqr. inch. Making use 



now of formula 53 we find the resultant unit tensile stress 
to be 



S't 



6,370 



+ V^^' 



548)^ + 



(6,370)' 



= 7,265 lbs. per 



sqr. inch, and from formula 54 the resultant unit shearing 
to be Ss = 



^(2,548)=^ + ^^'^^^^ = 4)08o lbs. per sqr. 



m. 



Prob. 139. In Fig. 74 prove that the diameter of the 
eye-bolt should be about one inch in order that the greatest 
unit tensile stress may not exceed 7,000 lbs. per sqr. inch. 

19 



136 



Art. 54. Combined Torsion and Flexure. 

The more important application of equations 53 and 54 
is to shafting, which in addition to transmitting power, 
must also sustain loads such as pulleys, fiy- wheels, etc., 
and the bending action due to belts. In this case the unit 
tensile or compressive stresses due to bending should be 
determined, as in Chapter VI, by the equation St =■ M c -=- I 
and the unit shearing stress due to the twisting, as in Chap- 
ter VIII, by the equation Ss = P 1 c -^ J. 

These values substituted in the above general equations 
will give us the resultant or maximum values required. 

If the horse power and number of revolutions per minute 
are given, we may make use of equation 49, from which 

Ss = — ^ ^^ Generally a shaft is round, in which 

JN 

c '^2 c 16 

case — =- = -^^<5 and -^^ =z -„ The equations for 

I TTd J Tta^ ^ 

resultant tension and shear therefore become 



TTd 7td\ N- 



e/ _ 16 / (63,ooo)-H^" 2 

^ ^ - "^^V N^ + 

Solving these equations for d^ we find 



'' = 'S + ^^l'^''r^ + ^' t^^^ 



d' = _i|_^ (63.cgorH' ^ j^j. ^^^^ 

From these the proper diameter can be calculated when 



137 

the bending moment, horse power to be transmitted, speed 
of shaft, and safe unit stresses are knovv^n, the greatest value 
of d being of course adopted. 

For example, a steel shaft is to transmit 50 H. P. at a 
speed of 100 N, the factor of safety being 10. The distance 
between bearings is to be 8 feet, and a pulley to be placed 
midway will weigh with the vertical belt tension upon it, 
1,000 lbs. The greatest bending moment will be at the 
center, and, neglecting the weight of the shaft and consid- 
ering it as a simple beam 

-. Wl 1,000 X 8 X 12 ir. • 1 

M = = -=: 24,000 lbs. inches. 

4 4 

St = 71,000 -f- 10 = 7,100 lbs. per sqr. inch. 
Ss = 57,000 ^- 10 = 5,700 lbs. per sqr. inch. 

Substituting these and the other given values in the 
above equations we find that in order to provide for the 
greatest tension 



16X24,000 ^ 16 /(63^or(5o)^_j_^^^^^^^. 



3.i4X7>ioo 3.14X7.100^ (100) 
d^ ■=. 17.2 + 24.2 = 41.4 
and d =. 3.46 inches. 
To provide for the greatest shear 



16 ^ /(63,ooo)^X(5o) 



V(63,ooo) X(5o) , , .2 

, V2 + (24,000)' = 30.1 

^ . . (100)' I V ^» / 

or d = 1^30.1 =3.1 inches. 

The diameter of the shaft should therefore be 3.5 inches 
in order to provide for the tensile stress produced by the 



38 



pulley. Leaving this stress entirel}- out of account, and 
using the formula for round shafts, Art. 47, 

d = 68.5V—— = 68. 5a/ ^ = 3.04 inches, 

A bs N _ \ 5,700X100 

from which it will be seen that the pulley is a material 
factor in the problem. 

Prob. 140. Find the diameter of a soft steel crank-shaft 
w^hich in addition to transmitting 100 H. P. at a speed of 
200 N, sustains also a fly-wheel weighing 2 tons, at a point 
2 feet from the crank bearing and 3 feet from the outer bear- 
ing, the factor of safet}^ being 10. Ans. d =: 4.44 inches. 

Prob. 141. On a 2%^^ shaft transmitting 30 H. P. at a 
speed of 150 N, there are two pullej^s ; one weighing 500 
lbs. is 10'' from a bearing, and the other weighing 100 lbs. 
is half way between the bearings which are 8 feet apart. 
Find the greatest tensile and shearing stresses. 

Ans. St = 6,124 Ss = 4,442. 

Prob. 142. A 3'' soft steel shaft transmits 80 H. P. at a 
speed of 150 N. The distance between two adjacent bear- 
ings being 6 feet, what load applied at the center would 
stress the material to the elastic limit? Ans. 4,420 lbs. 



Art. 55. Combined Torsion and Compression. 

Under this heading may be included screw propeller 
shafts and vertical shafts which in addition to transmitting 
power must sustain heavy loads. 

The unit shearing stresses produced by torsion ma}^ be 



139 

determined as in the last article, and the compressive stress, 
if the distance between bearings is small, by the formula 
Sc = P ^ A. 

Generally, however, it is necessary to employ Rankine's 
formula for long columns, given in Art. 48. 

Thus for torsion 

63,000 cH 16 X 63,000 XH 

and for compression, 



A \ r / 71: d 

These should be calculated separately and substituted 
in equations 53 and 54. But if the diameter is the quan- 
tity required, the equation becomes too complicated for 
algebraic solution and resource should be had to the 
graphic method alreadj^ described. Thus let a steel shaft 
transmit 50 H at a speed of 100 N, and at the same time 
sustain an end load or thrust of 6,000 lbs., the unrestrained 
length being 8 feet and the factor of safety 10. Let us 
assume first a diameter 2, then 3 and finally 4 inches and 
caculate the corresponding value of Ss and Sc from the 
equations above. Doing this w^e find that when 

d = 2'', Ss = 20,103 and Sc = 4,732 lbs. per sqr. inch. 

d = 3^', Ss =■ 5,945 and Sc = 1,406 lbs. per sqr. inch. 

d = 4^', Ss = 2,508 and Sc = 653 lbs. per sqr. inch. 

Substituting these values of Ss and Sc in equation 53 we 
find for the corresponding values of the resultant compres- 
sive unit stress, when 

d = 2'', S'c = 22,607 l^s. per sqr. inch. 

d = 3''', S'c = 6,690 lbs. per sqr. inch. 

d = 4'', S'c = 2,856 lbs. per sqr. inch. 



140 

Using plotting paper and with these values of d and S'c 
draw next a curve showing how S^c varies with d, as de- 
scribed in Art. 51. The ultimate strength of steel under 
compression is 80,000 lbs. per sqr. inch and the allowable 
w^orking stress in this case is 8,000 lbs. per sqr. inch. 
I^aying off to the same scale vS'c = 8, coo, we find the corre- 
sponding value of d to be about 2.82 inches. 

Next substituting the same values of Ss and Sc in equa- 
tion 54, w^e find for the corresponding values of the result- 
ant unit shearing stress, 

when d r:= 2, S's = 20,241 lbs. per sqr. inch, 
when d = 3, S's = 5*987 lbs. per sqr. inch, 
when d r= 4, S's = 2,550 lbs. per sqr. inch. 

Using these values, construct another curve, showing 
now the relation between d and Ss . The greatest allov/- 
able unit shearing stress in the shaft is 57,000 -=- 10 = 
5,700 lbs. per sqr. inch. Laying this off on our plotting 
paper to the same scale, we see that the corresponding 
value of d is about 3.15 inches. 

Hence we see that in order to provide for the compressive 
stress caused by the loads, the shaft should be 2.82 inches, 
while for the shearing stress it should be 3.15 inches, the 
latter is therefore the proper diameter to adopt. 

Prob. 143. A 4'' vertical steel shaft in addition to trans- 
mitting 100 H at a speed of 200 N, sustains a load of 10 tons 
placed at a distance of 8 ft. from the step bearing upon 
which the shaft rests. What is the greatest unit stress in 
the shaft? Ans. 3,820 lbs. 

Prob. 144. The engines of a steam-ship develop 5,000 
H and give 120 N to the screw, the speed of the ship 



141 

being 20 miles per hour. Assuming the efficiency of the 
propeller to be 50 per cent, what is its thrust upon the 
shaft? Ans. 46,880 lbs. 

Prob. 145. What, in Prob. 144, should be the diameter 
of a soft steel shaft, disregarding the thrust, and assum- 
ing a factor of safety of 10 ? Ans. 12.3 inches. 

Prob. 146. What should be the diameter taking into ac- 
count the thrust and assuming the distance between bear- 
ings to be so small that the shaft may be considered as a 
short column? Ans. 13.28 inches. 



CHAPTER X. 
STRESSES IN ROOF AND BRIDGE TRUSSES. 



Art. 



Definitions. 




A truss is a framed structure so arranged that the stress 
caused by external forces in each member acts in the 

direction of its 
length, or is 
either tensile or 
compressive. 
Thus Fig. 75 
Rj^- {j-^' [ fR^ represents a sim- 

K a i ' pie roof-truss 

^ — 1 —^ formed of three 

Fig. 75. pieces and resting 

upon the supports A and C. A load P applied at B would 
cause a compressive stress in the rafters AB and BC, and 
a tensile stress in the tie-rod AC. Of course the weight of 
the rafters or of an}- load resting upon them causes also 
bending or flexural stresses. These can be determined as 
in Chapter VI, but in what follows, the loads will be as- 
sumed as concentrated at the joints, and the tensile and 
compressive stresses caused by them will alone be consid- 
ered. If loads are applied at other points, as for example 
a weight of Pi pounds at O, the proper proportional part of 
Pi will be taken as acting at B and the remainder at A. 



143 

Thus if Pi is midway, one half acts at B and the other 
half at A ; or if at one quarter of the distance from B to A, 
then % of the weight acts at B and % at A. 

A uniform load, as in Chapter V, will be considered as 
concentrated at its center of gravity. 

Art. 57. Reactions, Moments, Verticai, Shear. 

Reactions should be determined by the method of mo- 
ments as in Chapter V. Thus, in the above Fig., the 
reactions Ri and Rj may be found from the equations of 
equilibrium^ 

RJ = Pa + Pib or R, = ^ ^ "^ ^'^ 

Ril = P(l — a) -f Pi(l — b) or 

J, ^ P(l — a) + Pi(l — b) 

and Ri -f R2 = P + Pi 
The moment of external forces about any point should 
be determined as in Chapter VI. Thus, the moment of all 
the external forces to the left of the apex B about this 

point M = Ria — Pi(a — b) :. .: (a) 

and of those to the right about the same point 

M = — R^Cl — a) -: ...: (b) 

These moments must be numerically equal because the 
sum of the moments tending to turn the truss in one direc- 
tion about any point, must exactly equal the sum of those 
tending to turn it in the other. Substituting the values of 
Ri and Rg already given, equation (a) becomes, 

^ ^ [P(l-a) + P.(l-b)]a _ ^^^^_^^ ^ 

Pla — Pa^ — P^ab+Pibl 
■^ 1 



20 



144 



and equation (b) becomes, 
P 1 a — P a' 



M = — 



P^ab + Pjbl 



1 



These are numericall}^ equal, the difference in the sign 
simply indicating that moments act in opposite directions. 

The vertical shear at any point should also be deter- 
mined as in Chapter V, being simply the difference between 
the sum of the upward and the sum of the downward forces 
to the left or right of this point. 

Thus, just to the left of B, 

V = Ri — Pi or V = R2 — P 
These values of V must be numerically equal, as maj- be 
readily proved b}^ substituting the expressions for Ri and 
R2 already given. 

Usually, for the sake of uniformit}^ moments and verti- 
cal shear of external forces to the left of the section under 
consideration will alone be determined, although some- 
times it is more convenient to use those to the right. 



Art. 58. Determination of Stresses. 




Fig. 76. 
is evidently one half the entire load 



Let us now con- 
sider the roof-truss 
ABCD, Fig 76, in 
which AD = CD 
and AB = CB, and 
assume that a total 
load of W pounds 
is uniformly distri- 
buted over each raf- 
ter. Each reaction 
, or R, = R, = W 



145 

pounds. One half tlie weight on each rafter may be taken 
as acting at B and the' other half at A or C. The down- 
ward forces are therefore W pounds at B, and ^W pounds 
at A and C. 

Draw any line XX, cutting the rafter, suspension-rod 
and tie-rod as shown in the Fig. The loads just found 
cause longitudinal stresses in these pieces. 

lyCt us now take away all 
that part of the truss to the 
right of the cutting line, leav- 
ing what is shown in Fig. 77, 
and replace the stresses by ex- 
ternal forces represented by A^ 
the arrows Li L, and L3. 

These forces must hold in 
equilibrium the loads W, Y^W 
and the reaction Ri ; and, since the truss does not move, 
the following important law can be stated. 

The moment of iiiternal stresses about any point must always 
equal the moment of the extertial forces which cause them. 

This is true whether, as in the Fig., the forces to the 
left of the cutting section are considered, or those to the 
right shown in Fig. 76. 

Let us now consider the stress Li acting along the rafter 
BC. Since the center of moments may be taken at any 
convenient point, let us place it where the suspension-rod 
BD and the tie-rod AD intersect at D. The moment of 
the stresses acting along these pieces about this point will 
evidently be zero. The moment of Li is 

M = Li X aD 

The moment of the external forces about the same point 

Ri 1 W 1 

M == — ~ 




146 

Placing these equal and solving for Li 
— 2 R^ 1 — W 1 
^' ~ 4 X aD 

To find the stress Lo acting along the tie-rod AD, choose 
the intersection of the other two pieces cut by the line XX 
as a center of moments, or the apex B. The resisting mo- 
ment of Lo about this point is M = L2 X eB. The exter- 

nal moment is M = — Placing: these 

2 4 

equal and solving for L2 , we find 

2Ril — Wl 



'& 



a.AlR.-^U^ 



4 X eB 
To find the stress L3 in the suspension-rod BD, prolong 
AD until it intersects BC at O, and make this point the 
center of moments. The resisting moment of L3 is M = 
L3 X Oa^. The moment of the external forces is 

M = Ri X a^A ^X a^A — wfa^A 

2 \ 2 

Placing these equal and solving for L3 

. ^' ^ Oa, 

It will be observed that we have found values of Li L2 
and lyg without indicating whether the stress is tensile or 
compressive. This can readily be determined by drawing 
an arrow along the piece cut by the line XX, and away 
from that portion of the truss to the left of this line. If, 
when the moment of the external forces is positive, the ar- 
row tends to produce a right-handed rotation about the 
center of moments for that piece, the stress is compressive; 
if a left-hand rotation, the piece is in tension. 



147 

If, however, the external moment is negative, that is, if 
the sum of the moments of the loads is greater than the 
moment of the reaction, the rule should be reversed. 

Thus, where XX cuts BC, drawing the arrow- as shown 
in Fig. 76, we see that it would produce a right-hand rota- 
tion about the center of moments D, hence Li is a com- 
pressive stress. An arrow" drawn along BD would produce 
left-hand rotation about the center of moments O. L3 is 
therefore a tensile stress ; and siniilarly the arrow along 
AD would cause left-hand rotation about B, hence L2 is 
also a tensile stress. 

The lever-arms Da, Be, Oa^, and Aag, can often be cal- 
culated ; or can be directl}'' measured from ^ parefully made 
drawing. Thus, if the angles a «fi and q> are known, 
and also the lengths of the pieces forming the truss, 
lever-arm Da = BD sin ^, and Be = BD cos oc^ 

The distance OD may be found from the proportion 

OD sin cp _ ^^ 

-DT, = -T-7 i : Oai = ODcosa'i 

and finally Aag = (AD -f- DO)cos<3ri 



Art. 59. Stresses in a Bridge Truss. 

Let Fig. 78 represent a simple triangular or "Warren" 
truss, having four panels, AF, FG, GH, and HI. The 
upper chord BB is parallel to the low^er chord AI, and 
the struts and ties joining these form equal isosoles trian- 
gles. We will designate the height of the truss by h, the 
span AI by 1, and the "apex loads" at F G and H by P, 
Pj and Pi2, lengths being in feet and forces in pounds. 



14-5 



The reaction Rj may be found by taking the center of 
moments at I. Thus, 

R,l = PX3/41 + PxX>^l + P^X^l or 

R, = -% P -f 14 Px + Ji P^ 

Draw the broken line XX, cutting the three pieces as 
shown, and first let us consider the stress in the upper 

n ^ \ B 03 C . ^ X D E 




or 

Fig. 78. 

chord CD, calling it L3 The other two pieces cut intersect 
each other at G, which, therefore, we will take as the center 
of moments. The moment of the stress in CD about, this 
point is M =r Lsh. The moment of the external forces 
to the left of XX about the same point is 

Since these moments must be equal 



L. 



_1_ 
h 



Ri 



P \ 



4 



i ( + ) 



To determine whether L3 is compressive or tensile, draw 
the arrow away from the cut as shown in the Fig. It 
would produce right-hand rotation about G, therefore the 
piece is in compression, usually designated by the sign -}-. 



149 

To determine the stress Lo in the lower chord FG, take 
C as the center of moments. The moment of L3 is 
M =: L5 h. The moment of the external forces is 

Placing these equal and solving for Iy5 

Drawing the arrow as shown, it would produce left-hand 
rotation about C. Hence the piece is in tension, usually 
designated by the sign — . 

We come nov/ to the stress Lg iti the diagonal CG. 
Since the top and bottom chords are parallel they will in- 
tersect only at an infinite distance. Let us assume O as, 
for example, a million miles from C. Designating by cp 
the angle the struts make with the chords, prolonging CG, 
and drawing OOi perpendicular to it, the moment of Lg 
about O is M = LpXOOi = LgXOCsin^^ 

The momemt of the external forces to the left of the cut- 
ting line is M = RiXOO^ — PXOO3 

Since O is at such a great distance, OO2 and OO3 are 
each practically equal to OC. 

Hence M = (Ri — P)X0C. Placing the moments 
equal LsXOCsin^ = (P-i - P)X0C. or 

I.a = ^'~^ (-) 

sm (p ^ 

But the numerator is the vertical shear for any point be- 
tween C and G. Hence we see that when the top and bottom 
chord or ' ^ flanges^ ' are parallel the stress in a diagonal or ' ^web 
Tneniber^'' may be found simply by dividi?ig the vertical shear by 
the sine of the angle which the diagonal makes with the chord. 



I50 

The character of the stress may be found as before by draw- 
ing the arrow as shown in the figure, and since it would 
produce left-hand rotation about O, the stress is tensile. 
If the point O had been to the left, the result would still 
be the same, for although our arrow would indicate a right- 
hand rotation, the moment of the external forces would be 
negative, R being greater than P, and the rule must be 
reversed. 

lyCt us now find the stress Li in the end strut AB. Draw 

V 

the cutting Ime YY. By the rule lust found, Li = ^ 

° -^ ^ sm cp 

Ri 



( + ) arid since the arrow Vv^ll produce right-hand 

sm qj ^ ' ^ '^ 

rotation about O, the piece is in compression. To find the 
stress 1^4 in the chord AF, take B as the center of moments; 

then 1^4 X h -= Ri X 1 -^- 8, or L4 = -^ (— ) which the 

o n 

arrow shows to be tensile. In a similar manner the stresses 

in all the other members of the truss can be found. 



Art. 60. Dkad Loads. Roof Trusses. 

The stresses occurtiiig in roof and bridge trusses depend 
materially upon the weight of the structure, and conversely, 
the weight depends u^Don the stress. It becomes necessary 
therefore, to assume a probable weight for the truss, and 
then knowing what additional load it will be called upon 
to bear, the stresses may be calculated. Having found 
these, the proper dimensions of each piece can be approxi- 
mately determined, and also the weight of the truss. This 
weight substituted in place of what was originally assumed, 



151 



the work of calculation may be repeated if desirable. From 
a discussion of results by Ricker in "Construction of Truss- 
ed Roofs," Prof. Merriman finds"^^ the following empirical 
formulae to be approximatel}^ correct: 

For wooden roof trusses, 



\V = 



al ( 



ii + 



lO 



For iron roof trusses. 



W 



3^1 

4 



+ 



[57] 



In which W is the weight of the truss in pounds, 1 its 
length in feet, and a the distance between adjacent trusses. 
Thus in a wooden truss of 100 feet span, with 15 feet be- 
tween trusses, the weight of each truss would probably be 



about 



I•■'^ 



100 



+ 



100^ 



'2 \ ■ 10 

iron truss would be likely to weigh 



= 8,250 pounds; while an 



3 X 15 X 100 



+ 



ICO 



12,375 pounds. In order to 



4 \ ■ 10 

find the total dead load acting upon a roof truss, to the 
truss itself must be added the weight of the roof covering, 
and, if the slope is less than 60 degrees, of snow lying upon 
it. The weight of this covering, consisting of "purlins" 
or cross beams, sheathing, shingles or slate, etc., should be 
calculated in each case, and will vary from 5 to 35 pounds 
per sqr. foot. The weight of snow vvull depend on location, 
varying from about 30 pounds in Canada to about 20 pounds 
per sqr. foot in New York. Thus in the above example, 
assuming the rafter to have a slope of 30 degrees, its length 

* Roof aud Bridge Trusses, Part T, p. 3. 
21 



152 

is 50 ~ CCS 50° == 57.7 feet; and if v^-e take a covering 
load of 20 lbs. per sqr. foot, and a snow load of 20 pounds 
additional, tlie total load resting on the truss would be 

2X-|oX(57.7X 15) = 69,200 pounds. 

x^dding this to the weight of the iron truss itself, we find 
for the grand total a weight of 81,400 pounds, which must 
be supported b}- the side walls or columns, each reaction 
being 40,700 pounds. 

To the stresses produced by this loading in the members 
constituting the truss, should be added those caused by 
the wind, as explained in another article. 



Aet. 61. Dead i^oads. Highway Bridges. 

The dead load acting upon a bridge truss is due to the 
weight of the truss itself and of the ficor s^'stem, consisting 
of planking, stringers, paving, etc., resting upon it. This 
will var}' greatl}-. From a discussion of actual weights 
given in Waddells' ''Highwaj- Bridges," Merriman gives 
the following formula, 

w = 140 -|- 12 b -|- 0.2 bl — 0.4 1 [58] 

in which w is the total load in pounds per linear foot, b 
the width of the bridge in feet, including sidewalks if any, 
and 1 the length of the span in feet ; one half of this 
weight being supported b}' each truss. 

The weight of iron and wooden bridges will not var}' 
greatly when of equal strength. 

Thus let the span be 100 feet, the width of the roadway 
be 15 feet, and of each overhanging sidewalk be 5 feet. 
The entire vridth is therefore 25 feet. Substituting in the 



153 

above eqnation, we find the probable dead load to be w = 
900 pounds per linear foot, or a total of 90,000 pounds. 

Each truss must support one half of this, or about 45,000 
pounds, besides whatever "live" or rolling load may come 
upon it. 

Art. 62. Dkad Loads. Raii^road Bridges. 

The approxim^ate dead load of an iron railroad bridge, 
including rails, cross ties, stringers, floor beams, and lat- 
eral bracing, may be found from the following equations 
given by Merriman. 

For single track \y = 560 -j- 5-61 ) 

>- [59] 

For double track w ==: 1,070 -}- 10.7 1 \ 

in which w is the dead load in pounds per linear foot, and 
1 the span in feet ; each truss supporting one half of this 
weight. 

Thus, if the span is 200 feet, a single track bridge is 
likely to weigh not very far from 

w = 560 -f 5.6X200 = i,6So lbs. per linear foot, or a 
total of 336,000 pounds ; each truss supporting about 168- 
000 pounds. It should be remembered that the form^ula 
gives simply an approximate weight which may be used in 
a preliminary calculation, and which should afterv^-ards be 
revised. 

Art. 63. Practical Applications. 

lyet Fig. 79 represent a Vv'ooden roof truss in which the 
span is 100 feet and the inclination of the rafters 30 de- 
grees, the chord AC and rafters AB and BC being divided 



154 

into equal parts by the tie-rods and struts ; and assume the 
trusses to be 15 feet apart and to sustain a combined dead 
and snoAv load of 40 lbs. per sqr. foot. The weight of each 
truss, b}' formula 57, is likel}- to be about 8,250 pounds. 



19300 



193(X) [^ :B 




Each rafter is 50 ^- 00330° ^ 57.7 feet long, and must 
support 40X57-7X15 = 34,600 pounds. The total load is 
therefore 8,250 -r- 2X34,600 = 77,450 pounds. Assuming 
this to rest uniformly on the rafters, the division AD 
would support 34^ 77,450 = 19,300 pounds (nearl}^), which 
considered as concentrated at the ends would cause a 
downward pressure of j/2 19,300 = 9,650 pounds at A and 
at D. Treating the other divisions similarly, we get the 
results shown in Fig. 79. Each reaction v/ill be 
Ri= Ro = % 77,450 = 38,700 pounds. 
To find the stress E3 in the piece AD, draw a cutting 
line XX, and take the center of moments at F. Any other 
point could have been chosen, but F is convenient since 
the distance AF is known. The moment of E3 is, 

M = E.XFD (FD being perpendicular to AD.) 



155 

The moment of the external forces to the left of the cut- 
ting line is M = RiXAF — 9,65oXAF 

Placing these equal, making FD = AF sin 30°, and solv- 
ing for Iv3 

AF(38,7co — 9,650) , ^ 

L3 = ^^ ;L. . V ^ = +57.900 PO^^ds 

Ai^ sm 30 

the arrow indicating compression. 

To find the stress Li in AF, take D as the center of mo- 
ments. Then 

L.XDoi =: PviXAoi — 9,65cXAoi = 28,95oXAoi 

But Aoi = 25 feet, and Doi = Aoitan3o° = 13.85 feet. 

25X28,950 

Hence h^ = — —^^ — = — 52,265 pounds. 

I3-S5 

the arrow indicating tension. 

To find the stress in DB, draw the cutting line YY, and 

take the center of moments at F where the other two pieces 

cut intersect. Then 

I.,XFD = (Ri — 9,650) XAF — i9,3ooxFoi 

804,330 , ^ ^ 

or L4 = = + 48,260 pounds. 

33-3X0.5 

To find the stress in BF, take A as the center of mo- 
ments. Then 

IvgXAoo = i9,30oXAoi = 482,500 lbs. feet. 
The lever-arm A02 might be measured with sufficient 
accurac}' from a carefull)^ made drawing, or can be found 
trigonometrically. Thus in the triangle ABF, 

A -r^ SO so 

AB = ^ — , = -~rr- = 57-7 feet, 
cos 30 0.866 

AF = 33.33 feet, and the included angle is 30° Calling 
the other angles B and F, 



153 

tan M(F - B) = 4|-=4gcot-^ = ^1-4X3.73 ^ ^ 

AB -p .-^r 2 91.06 

Hence >^^(F — B) :=r 45° or F — E = 90°. But F + 

B =r i8g° — 30° = 150°. Eliminating B from these two 

equations, we find F = 120°, and B = 30°, 

The lever-arm Aoo can now be readil}- be found, for 

Aoo = ABsinB = 57-7X0.5 = 2S. S5 feet. Hence 

_ 4^2,500 

-Lr, — ;t-t = — 10,710 pounas. 

20.<55 

To find the stress L. in the rod FG, take the center of 
moments at B. Then 
L2XABsin3o° = (Ri — 9,650)50 — 19,300X25 = 965,000. 

965,000 
^- ^- — ^Q Q- — = —33,450 pounds. 

25. bj 

Finalh^ to get the stress in DF, draw a cutting line and 
take the center of moments at the point of intersection 
of the other two pieces produced if necessary as at A. 

Then L., X AD = 19,300 X AOi = 482,500 lbs. feet. 

_ 482,500 , ^ . 

or L, = — -^ — = -p 16,710 pounds. 
2b. b5 

Since the truss is symetrical and loaded uniforml}^ the 
stresses in the right-hand portion are the same. We have 
therefore found all the stresses due to the dead loads. To 
these should be added any additional stress caused by v:ind, 
traveling cranes, or other forces. The proper dimensions 
can then be found as explained in the preceding chapters, 
proper attention being given to stresses caused by flexure 
or bending. 

Art. 64. Stresses Due to AVixd Pressure. 
In small roof trusses it is not usuallv necessary to consider 



157 

stresses produced by the wind, but in these presenting a 
large expanse of surface such stresses should be carefully 
calculated and provided for. Let Fig. 80 represent a roof 
truss acted upon by a normal pressure of n pounds on each 




Fig. 80 



square foot. If 1 is the length of the rafters and d the dis- 
tance between them, the total normal pressure is 

N r= n 1 d pounds. 

In the figure one half of this may be supposed to act at 
the apex D and one fourth at A and B. The reactions 
caused by these forces must be directl}^ opposed and may 
be found in the usual way, thus 

N 



R2 X AH 



4 

or 



X AB 



R. =. 



In the present case, 2 AD 
AH = AC cos (p 

N 



N N 

H X AD = - (AB + 2 AD) 

2 4 

N (AB + 2 AD) 
4 AH 
AB = AC -^- 2 cos 9 and 



Hence Ro = 



4 cos (p 



and Ri =r N -- 



N 
4 cos^ (p 



158 

The stresses due to the wind may now be found by the 
method of moments as in the preceding examples. Draw 
the cutting line XX. To find the stress in DB, take the 
center of moments at F, then 

h\ X DF z= ; R, — ^ XAD or L/ - -L^t. -^ ^"^-^^ 

\ 4 / 

And similarly- , 
L/ X Ao, = %XxAD 



>.iJL,' V^J. JL_,^ 


1 : -^^1 

\ 


4, DF 


or L/ = 


/2 -N X 


AD 
Ao, 


— 'XAB - 

47 


X AB 
— X — 

2 2 




R, — ^X 







L/ X AB sin cp = Ri — 

or L./ = — 

sm (p 

The stresses in the other pieces ma}^ be found similarly. 

The stresses in the right-hand members will not in this 
case be the same. Thus, drawing the cutting line YY, we 
may consider the forces to the left as hitherto, or for sim- 
plicity, in this instance, take onl}- those to the right of this 
line. 

For the rafter CE take the center of moments at G, then 

L' X EG r= R. X G03 or E' = 4- R, X §^ 

Our arrow in this case should be drawn away from the 
right-hand portion of the truss as shown in the Fig., and 
would indicate tension were not the moment of the reaction 
negative also, and since this reverses the rule, E' is a com- 
pressive stress; but since the wind is as likely to blovv' on 
one side as the other, there is no need of finding these 
stresses, those in the left-hand portion being much greater. 

Art. 65. PrEvSsure Caused by Wind. 
The pressure caused by wind blovring upon a fiat surface 



159 

held at right angles to it as determined by experiments by 
Prof. Langley and others, may approximately be found 
from the equation 

P -= 0.00357 V' [60] 

in which P is the pressure in lbs. per sqr. foot, and V is 
the velocity of the wind in miles per hour. 

The values of P thus found are somev/hat less than 
those indicated by theory. Thus, at a velocity of 100 miles 
per hour, the recorded pressure was 35.7 lbs. while theory 
would call for 53.8 lbs. It is customary, however, in design- 
ing buildings, to assume a possibility of the pressure reach- 
ing 50 lbs. per sqr. foot. 

The normal pressure caused by wind upon inclined 

surfaces cannot be found by sim^ple resolution of forces, 

experiment proving that for small angles it is much greater. 

The following table has been calculated from Duchemin's 

!N" 2 sin (p 

formula -r- =. , r—o — in which N is the normal pres- 

P I + sm (p ^ 

sure in lbs. per sqr. foot, and q) the angle of inclination of 
the surface to the wind. 



TABLE XV. 

__, 00000000 

95 = 5 10 15 20 25 30 35 40 

N 
— = 0.173 0.338 0.485 0.610 0.718 o.8co 0.863 0.910 

_-, O O 0/-0^0_0 OoO 

<P ^ 45 50 55 60 65 70 75 80 

N 
"^ = 0.943 0.967 0.980 0.990 0.995 0-999 0'999 I -OOO 

22 



i6o 

Thus, if the angle of a roof is 30 degrees, the normal 
wind pressure is 0.8 P, and if as above we assume 50 lbs. 
per sqr. foot as the greatest probable value of P, N would 
equal 40 lbs. per sqr. foot. Applying these results to Fig. 
80, the total normal pressure upon the rafter AB is 

N = 57-7X15X40 = 34»62o pounds. 

This, substituted in the formula for the reaction, gives 

_ 34,620 

Ri = 34.620 — = 23,080 pounds. 

4X(o.b66) 

The stresses due to the v/ind pressure are therefore, 

L/xDoi = (Ri — i<N)xAD = 14,425X28.85 

^ , 14,425X28.85 
or 1^1 = ■ = — 30,050 pounds. 

, 23,080 — Y-z 34,620 
L2 =^ = — 11,540 pounds. 

, 14,425X28.85 
L3 = —77 — + 24,970 pounds. 

ID. 00 
, 14,425X28.85 

L4 — r .-r — + 24.970 pounds. 

10.06 

^ , 17,310X28.85 , 

W = ' 00- ^ = +17.310 pounds. 



28. 8 



i> 



^ , 17,310X28.85 - 

^' = Ao.. = 28.85 = --17,310 pounds. 

These stresses should be added to those found in the 
preceding article for dead loads, in order to get the 
greatest in each case. 

Art. 66. Stresses Due to a Single Moving IvOad. 

Let us assume our roof-truss, in addition to supporting 
its own weight, the covering, and the pressure due to the 



i6i 



vvind, carries also a traveling crane which may be moved 
into any position on the lower chord as shown in Fig. 8i. 
The stresses produced in the members will evidently 
depend upon this position. 







X ^0, 



P 



FiCx. Si, 



Designating the moving load b}^ P, let us assume it at a 
distance of ai feet from the apex F and a.^ feet from G. 
Designating FG = AF =z CG by m, the portion trans- 
mitted to F will be Pi = V a^ -^ m and to G, P2 = P ai -^- m. 

2 m Pi -f m P2 2 Pi + Po . 



The reaction Ri =z 



3 m 



or, substituting values of Pi and P2 and reducing, 



Rx = 



P(2 a2 + aj 
3m 



Draw the cutting line XX. For the piece DB, the cen- 
ter of moments is at F, and calling the stress S4 we have 

, Ri m 



S.XFD = RiXAF or S, 



FD 



Evidently S4 will be greatest when Ri is greatest, and 
this will occur when the load P is at the vertex F, or when 
ai = o, and aj ■= m, making Rj =. YzV. Substituting 



l62 
IV /r • r^ , 2 P m 

Maximum 84 = -^ — -— pounds. 

' 3 FD ^ 

If the load is carried to the left of F, the line XX may 
be made to pass to the right of it, and in finding the result- 
ant moment of the external forces about F, we should need 
to subtract the moment of P from the moment of Rj. Thus 
suppose P to be midway between A and F. The reaction 

S P 
would then be Ri r:: -— - and our equation becomes 

6 

_,^^ 5P ^,m Pm ^ Pm 

S4X DF = V-Xm— PX- — or S^ = 



6 2 3 ' 3 X DF 

This value of S4 is only one half as great as when P was 

at F. Hence we see that the greatest stress in DB occurs 

when the load is at the apex just to the right of the cutting 

line. 

For the diagonal DF, the center of moments is at A, and 
the moment of the stress is S5 X AD, but since XX may 
always be drawn to the left of the load there is no external 
force tending to produce a stress in this piece. Hence Sg 
=z o. The center of moments for AF is at D, and 

^ ^ ^ A 2 P X Aoi 

Si X Do, = Rx X Aoi =1 ^ '- 

3 

when P is at F. Let us now suppose P to move to a point 

midwa}^ between A and F. One half of P may be supposed 

to act downward at F and the other half at A. The total 

reaction is five sixth's of P and our equation for S becomes 

^V P P 

Si X Doi = ^ X Aoi — - X Aoi = - X Ao^ 

P Aoi 
Hence for this position S, = — X -^ — or only one 

half as srreat as when the entire load was at F. 



i63 

It is difficult to make a rule appl3dng to all cases, but 
generally the stress in any piece v/ill be greatest when the 
load is placed at the first apex just to the right of the cut- 
ting line, when the left-hand portion of a truss is consid- 
ered. A study of the diagram, or a few trials will always 
enable the student to decide in each case. 

Assuming P to be lo tons, or 20,000 pounds, and other 
dimensions as in the preceding articles, let us now find 
the stress in each piece due to this load. 

2X2,000X25 

Si = ^^ = — 24,066 pounds. 

3X 13-^5 

6,666X50 -, 

02 = = — 12,030 pounds. 

27.7 

g^ ^ 13.333X33.33 ^ +26,660 pounds. 
16.66 

O4 = ;:— — = -j- 26,600 pounds. 

10.06 

S5 = . = 0.0 pounds. 

_ , 20,000X33-33 

®' + Ao, ^ 28.85 " -23,100 pounds. 

Collecting from the preceding articles the calculated 
stresses in the miembers comstituting this truss and ar- 
ranging in tabular form, 

AF FG AD DB DF BF 

Dead —52,265 —33.450 +57>900 +48,260 +16,710 —16,710 

Wind —30,050 —11,540 +24,970 +24,970 +17,310 —17,310 

P — 24,066 — 12,030 +26,660 +26,960 +co,ooo — 23,100 

Total 106,381 57,020 109,530 99,890 34,020 57,120 

_ _ + + + __ 



164 

Knowing now the total stress each piece will be called 
upon to bear, the necet'sary dimensions can be determined 
as explained in the preceding chapters. 



Art. 67. Stresses Due to Live Loads. 

Let Fig. S2 represent a highway or railroad bridge, 
which in addition to the constant weight upon it, may also 
be subjected to a moving or live load, such as a crowd of 
men, or train, which may cover the entire bridge or only a 
portion of it. 



X 



. a L. E 




Fig. 82, 



For convenience we will imagine the load to come on 
from the right. Let w^ equal the load per linear foot, 1 
the panel lengths AG, GH, etc., h the height AB of the 
truss, and q> the angle a diagonal makes ^vith the chords. 

Let us first suppose the load to extend from the vertex 
H to the right support, and consider the stresses in the 
top and bottom chords. Draw^ the cutting line XX. 



For CD the center of moments is at H and 

S. h = 2 R, 1 or Se = +-^^ 

h 

our arrow indicating compression. 

Next, let us suppose the load to come on from the left, 

extending from A to H, and causing a reaction R/. The 

2 R ' 1 

stress in CD would be Se' ==-" 4-— — r the arrow still 

n 

indicating compression. The total stress in this piece is 

therefore greatest when the load covers the entire bridge, 

and IS Se -h So == -. pounds. 

n 

Similarly, the total stress in the lov/er chord GH will be 
greatest when the entire bridge is loaded. Hence we may 
make the following important rule. 

T/ie greatest stress in the upper and loiver chords occurs 
when a uniform live load. cove7^s the entire span. 

Let us now consider the stress in the diagonal CH. By 
the rule deduced in Art. 59, the stress S^ = V -^ sin q->^ 
where V is the vertical shear between the points G and 
H. This shear equals V = Ri — any loads between A 
and H. If then the live load should extend to the left of 
H, we should have to subtract all that portion from the re- 
action in order to get the shear ; and since the increased 
reaction must be less than the load causing it, we may 
state another rule. 

The greatest stress in a diagonal occurs wheii the live load 
extends from the end. of the bridge to the apex nearest the 
cutti7ig line. 

Thus, if the load comes on from the right, it should ex- 
tend only to H ; if from the left, only as far as G. In the 



i66 

first case our arrow would indicate tension ; in the second, 
the vertical shear would be negative, and the arrow vrould 
indicate compression. We see therefore that a diagonal 
may be called upon to withstand stresses of an opposite 
character, depending upon which way the live load enters 
the bridge. 

The dead load, already considered in a previous article, 
would cause in CH a steady tensile stress which we will 
designate by Lio- A live load coming on from the right 
would cause an additional tensile stress of Sio pounds; and 
the total tensile stress will be the sum, or Lio -f Sjo lbs. 
We have just seen that a live load coming on from the left 
would cause compression in this piece, and designating it 
by S'lo, the resultant stress in CH is Lio — S'lo pounds. If 
S'lo should be greater than Lio, the rod must either act as a 
long column resisting compression, or some provision must 
be made to relieve it. This is usual I3' done b}- means of a 
counter-tie shown b}^ the dotted lines DG and DI in the 
Fig. The action of a load coming on from the left would 
be to relieve or take away entirely the tensile stress before 
existing in CH and to cause a tensile stress in the counter 
DG in place of a possible compressive stress in CH. Us- 
ually only the panels near the center of the span require 
counter-ties or braces. 

Art. 6S. I^ive Loads ox Highway Bridges. 

Highway bridges may at times be densely crowded with 
people, and also subjected to a succession of heavily loaded 
teams, the former being found to cause the greater stresses. 

The following table taken from Merriman's "Roofs and 
Bridges" gives values often adopted for this live load in 
pounds per sqr. foot. 



TABLE XVI. 



I.ive load. 


City 


bridges. 


Country bridges. 


Spans under 50 feet 




100 


90 


vSpans 50 to 125 feet 




90 


80 


Spans 125 to 200 feet 




80 


70 


Spans over 200 feet 




70 


60 



Thus in a city bridge 25 feet wide and spanning 100 feet, 
the greatest live load per linear foot ma}^ be taken as 25 X 
90 = 2,250 pounds, each truss supporting one half, or 1,125 
pounds. 

Art. 69. lyivK IvOADS on Ri^iLROAB Bridges. 

In a railroad bridge provision must be made for the 
heaviest locomotive and cars ever likel}^ to come upon it. 
A number of systems are used by engineers. One of the 
most simple is that adopted by Prof. Du Bois in his 
" Stresses in Framed Structures," in which a uniform load 
of 2,000 pounds per linear foot is assumed to represent the 
train, and in addition a "locomotive excess" or concen- 
trated load of 34,000 pounds to represent the extra v^^eight 
of a locomotive, one half of the total being carried by each 
truss. If there is more than one locomotive, Du Bois takes 
the distance apart of each "excess" as fifty feet. This 
S3^stem, which we wdll adopt, may be represented by the 
following diagram. In calculating stresses these "excess" 
weights should alw^ays be placed v/here they will do the 
most harm; i. e., 
cause the greatest 34^00 S'^OO 

stress Railroad Q . ^. ^'- ^'[^'"^ . ^ 

s.ress.^ Railroad __JTim^^ 

companies usually ! - t-. J 

specify the system ^ >M: rt ~~5, 

of loading wdien a Fig. 83. 

23 



1 68 



bridge is to be built, but the principles used in calculating 
stresses are the same in every case. 

Art. 7c. The Lexgth axd Depth of Panels. 

The length and depth var>- greatly, being often deter- 
mined by local considerations. The economic depth is 
that which makes the Aveight of material and cost of con- 
struction the smallest possible, and in general this will be 
found to varj' from one fifth to one eighth of the span. 
Assuming values for this depth and for the panel length, 
a few calculations will enable an engineer to determine 
which give best results. In practice the depth varies from 
one and four tenths to two times the length of the panel, 
which however seldom exceeds twenty-nve feet. Thus, if 
the span is to be 120 feet, we might divide the truss into 
eight panels, fifteen feet long and say twent}^ feet high. 

Art. 71- Application to a Pratt Truss. 



Let Fig. 84 represent a "through" Pratt iron truss for a 
single track railroad, the span being 120 feet divided into 




Fig. S-i. 



169 

8 panels each 15 feet long and 20 feet liigh. The probable 
weight of the bridge with its dead load, by x\rt. 62, is 

w = 560 -|- 5.6X120 = 1,232 lbs. per linear foot, 
one half of which goes to each truss. Bach lower apex 
therefore may be considered as loaded with 616X15 = 
9,240 pounds, and each reaction due to these loads is 
^X9, 240X7 = 32,340 pounds. Hence, in order to avoid 
the use of large figures, let us assume. 

Bach reaction Ri = R2 = 16 tons. 

Bach apex load P = 4.57 tons. 

The angle each diagonal makes with the chords is 

tan 9? = 20 -- 15 = 1.333 9 = 5S° 8' 
sin cp =: 0.8 cos (p = 0.6 

Stresses Due to Dead Loads. 

Designating the piece by the stress in it : 
For Li the center of moments is at a, and 

LiX 15 sin ^ == RiX 15 Li = -f 20 tons. 

For Iv2 the center of moments is at b, and 

L2X20 = R1X30 — RX 15 L2 = -h 20.5 tons. 

For Iv3 the center of moments is at c and 

L3X20 = R1X45 — ^(30 + 15) 1^3 = -h 25.7 tons. 
For 1^4 the center of moments is at d and 

1^4X20= R1X60— P(45 -f 30 + 15) L4 — +27.4 tons. 
For L5 the center of moments is at h and 

L5X2o = RiXi5 B5 = — 12 tons. 

For Le the center of moments is also at h and 

B6X2o=:RiXi5 Le = — 12 tons. 

For X7 the center of moments is at i and 

B7X20 = R1X30 — PX15 1/7 = -- 20.5 tons. 



For Ivs the center of moments is at j, and 

Iv8X2o = R1X45 — P(3o + 15) Ivs = —25.7 tons. 

For Lo the center of moments is at A, and 

1^9X15 = PXi5 L9 — — 46 tons. 

For the stress in the tie-rod hb, making use of the form- 
ula deduced in Art. 59, 

_ V _ R. - P ^ 

^'" - 11^' - sin cp L" = - ^4-3 tons. 

For the post bi, ^ = 90°, and 

T^ 2 P 

Ln '■= — —. ^ Lu = +6.9 tons. 

sm 90 ' 

Ri — 2 P 

-Uio = : Ivio = — 8.5 tons. 

sm cp 1- ^ 

Li3 = — -■ — ^o — Ivis = + 2.3 tons, 

sm 90 

j^ o p 

Li4 = — —■ — ^ Li4 = —2.9 tons. 

sm ^ ^ 

To determine the stress in the center post kd, draw a 
cutting line YY. This intersects four pieces, but since the 
stress in cd has already been found, our equation of equi- 
librium between internal stresses and external forces will 
contain only one unknown quantity. Thus taking j as 
the center of moments 

Ivi5 X 15 + Iv8 X 20 = Ri X 45 — P (30 + 15) 
But Ls X 20 = Ri X 45 — I" (30 + 15) (see above) 
Hence hi-. X 15 = o or Lis = o. 

There is therefore no stress in this piece due to dead 
load. Since the truss is symmetrical, we have now found 
all the stresses due to this loading and can consider 



lyi 



Stresses Due to Uniform Live Load. 

Adopting the system shown in Fig. 83, one half of the 
load being carried by each truss and considering first onlj^ 
the uniform train weight of 2,000 pounds per linear foot. 

Each apex load, P = i,ogo X 15 = 15,000 lbs. = 7.5 tons. 
And when the entire bridge is covered 

Kach reaction R = 7 X 7.5 h- 2 := 26.25 tons. 
It has been showm (Art. 67) that the greatest stresses in 
the top and bottom chords occur when the uniform live 
load covers the entire span. These ma}^ readily be found 
as before. Thus, designating corresponding stresses by 

Si X 15 sin ^ = R X 15 Si == + 32.8 tons. 

52 X 20 = R X 30 — P X 15 S2 = + 33-7 tons. 

53 X 20 = R X 45 — ? (30 + 15) S3 = + 42.1 tons. 

54 X 20 = R X 60 — P (45 + 30 H- 15) S, = + 45.0 tons. 
S- X 20 = R X 15 S5 = — 18.6 tons. 
Sg X 20 = R X 15 Sfi = — 18.6 tons. 
S: X 20 = R X 30 — P X 15 S, = — 33.7 tons. 
Sg X 20 == R X 45 — P (30 + 15) Sg = — 42.1 tons. 

To find the stresses in the posts and ties vv^e may make 
use of the rule that these are greatest when the load ex- 
tends from the end of the bridge to the apex nearest the 
cutting line. 

Since the tie-rod a h sustains only the load at the apex a 

S9 = 7-5 tons. 
For Sio the load extends from B to the apex b. In this 
case the moment of the reaction about B is 
RX8Xi5 = P(6Xi5 + 5X 15 + 4X15 + 3X15 

-h 2 X15+ 15) 



172 

Canceling the factor 15 and dividing b}^ S 

R zr: i/gP (6 + 5 -f 4 + 3 + 2 X i) = '^'^^^' = 19.7 tons. 

and since the load only extends to the vertex b, the verti- 
cal shear between A and b must also equal 19.7 tons. 

Hence Sio = -. = --^ S,o = — 24.6 tons. 

sm (p o.S 

For Sji the load cnlj^ extends to the vertex c because 
otherwise we should need to subtract loads at b or a in or- 
der to obtain the shear. 
R = /8 P (5 + 4 + 3 + 2 + i) == /8 X 7-5 X 15 = 14- 1 

-I rA V 14. I ^ 

tons, and bn =r ~ ^ =: S^ = -j- 14. i tons. 

sm 90 I 

For Sio the load also ends at c and 

Oio ::= — = bj. = — 17.6 tons. 

sm (p o.S 
For Si3 the load should end at d 
■R = :^8 P (4 + 3 -r 2 -f i) = ^ X 7-5 X 10 = 9.4 tons, 

and Si3 =: o Sj.; = -f- 9.4 tons. 

sm 90 

For Su the reaction and shear are the same 

o sin cu 04- 

Su = Si4 = — 1 1.7 tons. 

9.4 

As before shown S15 = o 

For the tie-rod dl the load should end at e 

R = }iF {3 + 2 -\- i) — Yz X 7-5 X 6 == 5.6 tons, and 

V R 5.6 o ■ ^ 

S16 = -• = -^ — -^ Sie = ^ 7-0 tons. 

sm (p sm qj 0.8 

It should be observed that in this case, our arrow indi- 
cates right-hand rotation or compression. A load coming 
on from the right would therefore cause compression of 7 
tons in the tie-rod dl v^^hile if it came on from the left it 
w^ould cause tension of 11.7 tons in the same piece. 



173 

For Sn the load should also end at e, and 

Sit = -^ -o Sn = — 5-6 tOnS. 

sm 90 

For Sis the load should end at f. 

R:=^P (2+ i) = 2.8 tons, and 

2.8 

Sm = —. Si8 = -h 3.5 tons. 

srn c/J ' ^ ^ 

2.8 

Siq == —. 5 S1.9 ==: — 2.8 tons. 

sm 90 

Soo = -^ — - Soo = + 1-2 tons, 

sm cp 

We have now found the stress in each member of the 

truss for the uniform live load and there onl}^ remains the 

Stresses Due to Locomotive Excess. 

The distance apart of each excess as before stated is 
assumed to be 50 feet. Thus if one or 17 tons were placed 
at the apex a, the other would fall at one third the distance 
between d and e. It is, however, usual in such cases to 
assume the second excess to act at the nearest apex, which 
in this instance would be at d, making the distance apart 
of these loads 45 instead of 50 feet. 

The problem is similar to that discussed in Art. 66 ; the 
general rule being to place the first excess under the center 
of moments for the top and bottom chords ; and at the first 
vertex to the right of the cutting line for posts and ties ; 
the position of the second excess of course being deter- 
mined by that of the first. 

Thus for the end strut Ae, place 17 tons at the center of 
moments a and 17 more tons at the'apex d. The reaction is 

R z= 2}i (7 + 4) = 23.4 tons, and designating corre- 
sponding stresses in the members by Tj, T2, etc. 



174 

Ti X 15 sin c/i' zrr R X 15 and T^ = + 29.2 tens. 

For Tn place the loads at b and e, then 
R :=z 17 (% -f- ^/s) = 19.1 tons, and T, =z ^ 28. 6 tons. 

For T3 place the loads at c and f 
R = ij (ys -\- }() = 14.9 tons, and T, =z -^ 33.5 tons. 

For T4 place the loads at d and g 
R = 17 0^ -L ^) — 10.6 tons, and T, = + 31.8 tons. 

For T5 the loads are again at a and d 
R rr= 23.4 tons, and T3 1= — 17.6 tons. 

For Tg the loading is the same as for T5 
R =. 23.4 tons, and Tg = — 17.6 tons. 

For T:, R r= 19. 1 tons, and T- = — 2S.6 tons. 

For Ts, R = 14.9 tons, and T^ = — 33.6 tons. 

For Tg, the stress is evidently ^0 =i — 17.0 tons. 

For the tie-rod bh, the loads should be at b and e. The 

reaction R. — ■ 19. i tons. 

ATA V R 19,1 ^ 

Tio = -. , = -. — - = — T- or Tio = — 23.9 tons. 

sm cp sm G^ 0.8 

For Til the loads should be at c and f because in finding 
the vertical shear for this piece any load at b must be sub- 
tracted. Therefore R r=: 14.9 tons, and 

Tn = 4^^^ Tn = + 14.9 tons. 

sm 90 

For Tio the loading is the same and 

Ti. — 4^^ T1.3 = - 18,5 tons, 

sm qj 

For Ti3 the loads being at d and g, R = 10.6 tons. 

Ti3 = ^^' o "Tis = + 10.6 tons, 

sm 90 

For Tii the loading is the same, 

AtA 10.6 tT\ J. 

Tu = -. T^i4 ■-= — 13-5 tons. 

sm fp 



T., 





1 


5- 


2 tons. 


Ti, 


= 


— 


4- 


2 tons. 


T.. 




+ 


2 


.6 tons. 



175 

As before Ti-, = c.o tons. 

For Tie the loads should be at e and B, R m 6.4 tons, 

Tie = -^ Tie = + 8.0 tons, 

sin cp 

It should be observed that in this and all following posi- 
tions, the stresses are of a different character to those in 
corresponding pieces on the left side of the truss. 

For Tn the loads being at e and B 

Ti: = . '"^ o Ti, = — 6.4 tons, 

sm 90 

For Tis, one load is at f, the other off the bridge. 

R 1= 4.2 tons. 

T — — 

sm 90 

^ 2.1 

T20 = -^—- 
sm cp 

We have nov/ found all the stresses produced by the 

dead and live loads in the members forming this bridge 

truss and are prepared to find the greatest stress each may 

be called upon to bear. 

TABUI.ATKD Results. 

Ah hi ij jk Aa ab 

Dead Load -|~40-o +4i-i ~l~5i-4 "l~54-8 — 24.0 — 24.0 

lyive Load +32.8 +33.7 +42.1 +45.0 — 18.6 — 18.6 

Lo. Excess +29.2 -|-28.6 +33.5 +31.8 — 17.6 — 17.6 

Total -(-102.0 +103.4 H-I27.0 -fi3i.6 — 60.2 — 60.2 

be cd ah bh bi ci 

Dead Load — 41. i — 51.4 — 9.1 — 28.6 -(-13.7 — 17. i 

Live Load — 33.7 — 42.1 —7.5 —24.6 +14. i — 17.6 

Lo. Excess — 28.6 — 33.5 — 17.0 — 23.9 -I-14.9 — 18.5 

Total —103.4 —127.0 —33-6 — 77-i +42.7 —53-2 

24 



1/6 





cj 


dj 


dl 


el 


em 


fill 


in 


Dd. Ivd. 


-f4.6 


-5.8 


-5.8 


+4.6 


17. 1 


+ 13.7 


—28.6 


Lv. W. 


+9-4 • 


II. 7 


+7.0 
II. 7 


-5-6 
+9-4 


+3-5 
— 17.6 


—2.8 
+ 14.1 


+ 1.2 
— 24.6 


Lo. Ex. 


+ I0.6 


— 13-3 


+8.0 
13-3 


—6.4 
+ 10.6 


+5.2 
—18.5 


—4.2 
+ 14.9 


+2.6 
--23.9 


Total 


+ 24,6 ■ 


—30.8 


—30.8 
+ 15.0 


-r24.G ■ 
— 12.0 


53-2 
+8.7 


+42.7 
7.0 


77.1 
+3.8 



From these figures the proper dimensions can be given 
to the several members. Thus the strut Ah should be 
treated as a long column designed to sustain an end thrust 
of 102 tons. The tie-rods bli and fm to sustain a pull of 
77.1 tons, the tie-rods dj and dl a pull of 30.8 tons. 

Counter Ties. 

It will be observed in the last columns that two sets of 
figures are given for the stresses due the live loads, the 
first being the results obtained supposing the load to come 
on from the right, while the second assumes it to come on 
from the left. Thus a live load extending from A to d 
would cause, together with the dead load, a total tensile 
of 30.8 tons in the tie-rod dl, or the same as that in dj were 
the load to extend from B to d. But when the live load 
extends from B to e, the table shows it would cause a com- 
pressive stress in dl of 15 tons, which must be provided for. 
A long column is necessarily heavier and more expensive 
than a rod in tension carrying the same load, and moreover 
as was seen in Chap. II, stresses varying from tension to 
compression, or the reverse, are much more injurious than 
those of the same nature. Therefore to avoid compression 



177 

in a long diagonal it is customary to insert counter ties 
shown by the dotted lines ke and kc. These must be pro- 
sportioned to sustain the difference between the compressive 
stress due to the live load and the tensile stress due to the 
steady load. Thus when a train is on the point of entering 
the bridge from the right, the rod dl sustains a tensile 
stress of 5.8 tons and the counter tie ke is unloaded. As 
the train proceeds this stress is taken off until in some po- 
sition none is left. Then the counter tie begins to take up 
tie burden which when greatest reaches a total of 15 — 5.8 
= 9.2 tons. The center post dk which hitherto has been 
useless now^ transmits this stress to the other members of 
the truss. The post el which under the dead load sustained 
a compressive stress of 4.6 tons nov/ becomes unloaded, the 
center post practically taking its place. 

The greatest stress in the center post dk nia3^ be found 
by simple resolution of forces. Thus along ke lay off a 
line to represent the stress of 9.2 tons and resolve it into 
components along the upper chord and post as shown in 
the figure. Designating by S15 the stress in dk we have 
Si3 = 9.2 sin ^^ = 9.2 X 0.8 or S15 = -f- 7.36 tons. 
The tensile stress in the diagonal em due to the dead load 
has been found to be 18. i tons, and the compressive stress 
due to live load to be 8.7 tons. The live load v/ould there- 
fore never entirely take away the tensile stress, leaving at 
all times as much as 17. i — 8.7 =: 8.4 tons. There is then 
no need of a counter in this panel. 

Art. 72. Double oii Multipi^k Systems. 

A system in which the tie-rods and struts are arranged 
in such a way that a cutting line cannot be drawn without 
i, k, etc.; and the members ah, he, etc., by the loads at the 



intersecting mere than tliree pieces is known as a multiple 
S3'stem. Thus Fig. 85 represents a double "Warren" truss 
and a cutting line XX will intersect at least four pieces. 
Other diagonals might be shown b}- the dotted lines on the 
right, forniing what is known as "lattice" truss or girder. 
In an}' case the stresses may be found in the same manner 




as before, when we consider that each rod or strut is stressed 
simpl}' b}^ the load applied at the apex where it intersects 
the chord. Thus if the load is on the bottom chord, the 
members Ab, bi, id, etc., are stressed b}- loads at the 



179 

apices h, j, etc. We may therefore take eacli sj^stem by 
itself as shown by the full lines in Figs. 86 and 87, and 
calculate the stresses as in the preceding article. 

For example let the figure represent a highway bridge, 
having a span of 120 feet divided into six bays 20 feet wide 
and 20 feet high, and let the width of the bridge be 30 feet. 

Dkad Load Strkssks. 

By Art. 61 the approximate weight will be w = 140 + 
12b -|- .2bl — .4I = 1,172 pounds per linear foot. Of this 
each truss will support one half or 586 pounds per linear 
foot. Kach apex load will therefore be 586 x 20 = 11,720 
pounds, or say P = 6 tons. 

Draw a line XX and let us find the stresses in all the 
pieces cut. The diagonal ib belongs to the system shown 
in Fig. 86, in which there are only two loaded apices, i and 
k. The reaction is therefore 

R zr: 34 (2 X 6) — 6 tons, 
The angle qj = 45°, and sin 9 = 0.707 
Designating the stresses by Li 
V R — 6 o 

JUi = -: = — : = -: Or hi = CO 

sm q) sm qj sm cp 

There is therefore no stress in this piece due to the dead 
load. To find the stress La in the diagonal cj belonging to 
the system shown in Fig. 87, R = ^ (3 X 6) = 9 tons, 

Ivo = —. = or Ivo = — 4.2 tons. 

sm cp 0.707 

To find the stress in cd and ij, we must combine those 
due to the apex loads on each system, since the chords are 
common to both. Thus from Fig. 86 the center of mo- 
ments being at i 

1/z X 20 = R X 40, or Iv'a = -f- 12 tons. 

25 



iSo 

From Fig. 87, the center of moments being at 3 

L^'s X 20 zr R X 60 — P X 20, V, = + 15 tons. 
The total stress in cd is therefore 

Ls = V, + V's = + 27 tons. 
Similar!}^ for the stress in the chord ij. 
From Fig. 86, L^ X 20 =z R X 60 — 6 X 20, or V, = 

— 12 tons. 

From Fig. 87, V\ X 20 == R X 40 — 6 X 20, or V\ = 

— 12 tons. 

The total stress in ij is therefore 

L, — V, — V\ = — 24 tons. 
In a similar manner the dead load stresses in all the other 
members mav be found. 



Stresses Due to Live Loads. 

Let us next find the stresses due to a possible live load 
of 90 lbs. per sqr. foot (See Table XVI). The width of the 
bridge being 30 feet, the load per linear foot would be 
2,700 pounds or 1,350 pounds on each truss. Each apex 
load is therefore 1,350 X 20 = 27,000 pounds, or P = 13.5 
tons. It has been shown that the greatest stresses in the 
web members occur when a uniform live load ends at the 
apex next preceding the cutting line. For id, in Fig. 86, 
this load vrould end at k, the reaction would be 

R = ^Xi3-5 = 4-5 tons. 
Designating live load stresses by Si, So etc., 

Sx = 4^^ == -^^^ S, = +6.3 tons, 

sm ^ 0.707 



I8i 

For cj ill Fig. 87, with the load ending at j, the reaction 
and shear would be 

Ri = ^3-5(-|- -r ^) ^ 9 tons. 

o _ 9 c 



02 = — 12.7 tons. 
0.707 

We have seen that the flange stresses are greatest when 
the uniform load covers the entire span. For the s^'steni 
shown in Fig. 86, the reaction would be 13.5 tons, and 
with the center of moments at i, 

83^X20 = RX40 S/ = + 27 tons. 

For the other system, R =^ i}4Xi3.5 = 20.25 tons. 

83^^X20 = 20.25X60— 13.5X40 ^/' = +33-75 tons. 

The total stress in the chord cd therefore is 

S3 = S/ -^ S/' — +60.7 tons. 
To find the total stress in the low^er chord ij, 
S/X20 = 13.5X60 — 13.5X20 8/ = —57 tons. 

8/^X20 = 20.25X40 — 13.5X20 8/' = — 26 tons. 
The total stress in the lower chord ij is therefore 

8, = 8/ + S/' = —83 tons. 
In a similar manner the maximum stresses due to the 
live load in all the other members may be determined. 

Tabulated Results. 

ab be cd Ah hi ij 

Dead I^oad +9.0 -{-21.0 -{-27.0 — 6.0 — 18.0 — 24.0 

lyive Load +i3-5 +47-2 +60.7 —13.5 — 40.5 — 83.0 

Total +22.5 +68.2 -i-87.7 —19-5 —58.5 —107.0 

Ab bi id dk kf Bf 

Dead Load +8.4 — 8.4 0.0 0.0 — ^8.4 +8.4 
Live Load -[-19.0 — 19.0 +6.3 — 6.3 — 19.0 -j-19.0 

Total +27.4 —27.4 +6.3 —6.3 —27.4 +27.4 



I«2 



Aa ah 

Dead I,oad +9.0 — 12.; 



he 
+4.2 



-4.2 



-4.2 



el 

+4 



Live Load -f-20.2 — 28.5 +12.7 — 12.7 



—12.7 +12 
+3-4 —3 



Total 



+ 29.2 — 41.2 -)-i6.9 



•16.9 



—16.9 +16 
+3-4 —3 

It will be seen that a load coming on the bridge from 
the left will cause a maximum compressive stress in the 
pieces je of 3.4 tons, and a maximum tensile stress in the 
piece el of 3.4 tons, whereas the dead load causes stress of 
an opposite character in each of these pieces of 4.2 tons. 
However, since the live load is less than the dead load 
stress these pieces will be relieved of part, and it will not 
be necessary to provide for stresses of an opposite nature 
in the same place. 

Art. 73. Thk Bowstring Girder. 

Fig. 88 represents what is known as a bowstring truss, a 
form very often used for highway bridges and roofs. The 
method of finding stresses is the same as in the preceding 




Fig. 



i83 

cases, although it is sometimes troublesome to calculate 
lever-arm.s. In such cases a carefully made drawing will 
generally determine these with sufficient accuracy. 

Let us assume the span to be loo feet divided into five 
equal bays, and the height in the center to be 30 feet, with 
the apices of the upper chord in an arc of a circle. The 
radius of this circle may readily" be found from the equation 

R' = 50' _)_ (R _ 30)^- z=: 2,500 + R' — 60R + 900 
or 60 R r= 3,400, from which R = 56.667 feet. 

Make an accurate drawing, using a scale of one quarter 
of an inch to the foot, and assuming the pieces Ab, be, cd, 
etc., as straight lines, find by construction, the lever-arms 
necessar}^ For example, supposing this highwa}^ bridge 
to be 25 feet wide, the total weight is likely to be 
W =■ 140 -|- 12 b -j- .2bl — .4I — - 900 lbs. per linear 
foot, or a dead apex load on each truss of j4 (900 X 20) 
=■ 9,000 pounds or 4.5 tons. The live load may be taken 
at 80 lbs. per sqr. foot, or 25 X 80 = 2,000 lbs. per linear 
foot, giving an apex load of j4 (2,000 X 20) = 20,000 lbs., 
or 10 tons. 

Draw a cutting line XX and let us find the greatest stress 
in the pieces be, bg, and fg. For be the center of moments 
is at g and its lever-arm og by measurement or calculation 
is 27 feet. Each reaction is R r= ^ (4 X 4.5) = 9 tons. 
Hence, L2 X 27 = 9 X 40 — 4.5 X 20, or L2 = + 10 tons. 

For bg the center of moments is where be and fg meet 
when prolonged at o and its lever-arm ooi is about 55 feet. 
The distance oA is about 35.4 feet and the distance fo 
about 55.4 feet. Hence, 

L., X 55 " — 9 X 35-4 + 4-5 X 55-4 U = — 1.3 tons. 

For fg the center of moments is at b and the lever-arm 
fb is 21.3 feet. 



iS4 

Hence, L.s X 21.3 =r 9 x 20, or Ls = — S.4 tons. 

Draw the cutting line YY. The center of moments of 
Ab is at f ; its lever-arm fo is about 14.6 feet, and 
Li X 14.6 = 9 X 20, or Li = -|- 12.3 tons. 

For bf the center of moments is at A, and 

L4 X 20 =: 4.5 X 20, or L^ = — 4.5 tons. 

In a similar manner, the maximum stress due to the live 
loads may be found, and the following table prepared, first 
considering the piece gd as left out. 





Ab 


be 




cd 


Af 


fg 


gli 


Dead load. 


-ki2.3 


-!- 10. 





1-9.3 


-8.4 


-8.4 


— 9-3 


Live Load. 


-t- 27.4 


-f 22. 


2 


— 20.7 


— 18. S 


— 18. 8 


— 18.8 


Total 


+ 39-7 


— 32. 


2 


-r 30.0 


27.2 


27.2 


30.0 




bf 


bg 




gc 


vJil 


hd 


he 


Dead Load 


— 4-5 


— 1-3 




— 0.4 


0.0 


— 0.4 


— 1-3 


Live Load 


— lO.O 


1-1 




+ 2.8 




-4.8 


-^ 6.7 






- 6.7 




-4.3 


1 / -o 


— 2.8 


7-7 


Total 


— 14-5 


9.0 




5-2 


7-3 


5-2 


9.0 






+ 6.7 




+ 2.8 


~r 7-3 


+ 2.8 


+ 6.7 



From this table it will be seen that without counter 
braces, the post and ties will be subject to both compressive 
and tensile stress. Thus the counter rod di should be capi- 
ble of bearing a tensile stress of 6.7 tons Vv'hich otherwise 
would fall upon the rod he causing compression. This 
counter will in turn cause a compression in the post dh of 
about 4.8 tons, (see table). Similarl}' the counter dg 
would take awa}' the compression of 7.3 tons caused by the 
live load in ch and transmitting it to the post eg cause a 
compression in it of about 4.8 tons, which is the greatest 
stress this post will be called upon to bear. 



i85 



Art. 74. Kinds of Roof and Bridge Trusses. 

A framed structure, as already pointed out, is essentially 
formed of a series of triangles joined together, this being 
the only figure which cannot be changed in shape without 
distorting the sides forming it, but these triangles can be 
arranged in a great variety of wa3^s. Accordingly there 
are many different arrangements in use. Some of the more 
common have already been illustrated. Thus Fig. 75 rep- 
resents the simplest form of all, a simple roof-truss most 
commonly used for 
short spans. When 
used for a bridge 
there is usuall}^ a 
center tie-rod or post 
as shown in Fig. 89, 
called a king-post 
truss, or the truss 
ma}^ be inverted as 
in Fig. 90, with the 
load on the upper 
chord instead of on 
the lower. When 
arranged as in Fig. 
91, it becomes what 
is known as a queen- 
post truss. 

Other forms of roof 
trusses are shown in 
figures 92, 93, 94. 
Many other arrange- 
ments might be made 
but the method for 




Fig. 



90. 



finding the stresses is 




Fig. 92. 



iS6 




generally tlie same. In some instances however, a cutting 
line cannot be dravv-n without intersecting four pieces, as for 
example, in finding the stresses in the pieces ab, ad, ae, or 

be in Fig. 93. But 
from the symmetry 
of the figure the 
stress in ab is obvi- 
ously the same as in 
cb, which can readi- 
l}'- be found, and ad 
is the same as dc. 
Since ab is known, 
the stress ae can be 
determined by tak- 
ing the center of 
moments at A, and 
so with other pieces. 
Fig. 95 represents 
a Howe truss in 
which the diagonals 
are in compression 
and the verticals in 
tension. Fig. 84 il- 
lustrates a Pratt 
truss, one of the most 
common forms in use 
and this ma}" also be 
inverted to form a 
"deck" bridge. Fig. 
97 represents a dou- 
ble Pratt or Whipple 
truss, also common- 
FiG. 97. ly used. 




Fi 



G. 95. 




Fig. 96, 




